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‎README.md

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@@ -175,7 +175,7 @@ python problem.py https://www.lintcode.com/problem/92 -l cpp
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## Simulation
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| Link | Problem(14) | Solution | Tag | Time | Space | Ref |
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| Link | Problem(15) | Solution | Tag | Time | Space | Ref |
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| ----- | ----- | ----- | ----- | ----- | ----- | ----- |
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| [Leetcode-67](https://leetcode.com/problems/add-binary/) | Add Binary | [c++](./leetcode/67.add-binary.cpp), [python3](./leetcode/67.add-binary.py) | Simulation | O\(M\+N\) | O\(1\) | - |
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| [Leetcode-415](https://leetcode.com/problems/add-strings/) | Add Strings | [c++](./leetcode/415.add-strings.cpp), [python3](./leetcode/415.add-strings.py) | Simulation | O\(N\) | O\(1\) | - |
@@ -184,6 +184,7 @@ python problem.py https://www.lintcode.com/problem/92 -l cpp
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| [Lintcode-849](https://www.lintcode.com/problem/basic-calculator-iii/) | Basic Calculator III | [c++](./lintcode/849.basic-calculator-iii.cpp), [python3](./lintcode/849.basic-calculator-iii.py) | Simulation | O\(N\) | O\(1\) | Leetcode-772 |
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| [Lintcode-553](https://www.lintcode.com/problem/bomb-enemy/) | Bomb Enemy | [c++](./lintcode/553.bomb-enemy.cpp), [python3](./lintcode/553.bomb-enemy.py) | Simulation | O\(NM\) | O\(NM\) | Leetcode-361 |
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| [Lintcode-366](https://www.lintcode.com/problem/fibonacci/) | Fibonacci | [c++](./lintcode/366.fibonacci.cpp), [python3](./lintcode/366.fibonacci.py) | Simulation | O\(N\) | O\(N\) | - |
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| [Lintcode-645](https://www.lintcode.com/problem/find-the-celebrity/) | Find The Celebrity | [c++](./lintcode/645.find-the-celebrity.cpp), [python3](./lintcode/645.find-the-celebrity.py) | Simulation | O\(N\) | O\(N\) | Leetcode-277 |
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| [Lintcode-22](https://www.lintcode.com/problem/flatten-list/) | Flatten List | [c++](./lintcode/22.flatten-list.cpp), [python3](./lintcode/22.flatten-list.py) | Simulation | O\(N\) | O\(N\) | - |
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| [Lintcode-640](https://www.lintcode.com/problem/one-edit-distance/) | One Edit Distance | [c++](./lintcode/640.one-edit-distance.cpp), [python3](./lintcode/640.one-edit-distance.py) | Simulation | O\(N\) | O\(1\) | Leetcode-161 |
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| [Lintcode-3622](https://www.lintcode.com/problem/read-n-characters-given-read4/) | Read N Characters Given Read4 | [c++](./lintcode/3622.read-n-characters-given-read4.cpp), [python3](./lintcode/3622.read-n-characters-given-read4.py) | Simulation | O\(N\) | O\(1\) | Leetcode-157 |
@@ -925,7 +926,7 @@ python problem.py https://www.lintcode.com/problem/92 -l cpp
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## Other
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| Link | Problem(124) | Solution | Tag | Time | Space | Ref |
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| Link | Problem(123) | Solution | Tag | Time | Space | Ref |
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| ----- | ----- | ----- | ----- | ----- | ----- | ----- |
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| [Leetcode-16](https://leetcode.com/problems/3sum-closest/) | 3Sum Closest | [c++](./leetcode/16.3sum-closest.cpp), [python3](./leetcode/16.3sum-closest.py) | Other | \- | \- | - |
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| [Leetcode-454](https://leetcode.com/problems/4sum-ii/) | 4Sum II | [c++](./leetcode/454.4sum-ii.cpp) | Other | \- | \- | - |
@@ -960,7 +961,6 @@ python problem.py https://www.lintcode.com/problem/92 -l cpp
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| [Leetcode-652](https://leetcode.com/problems/find-duplicate-subtrees/) | Find Duplicate Subtrees | [c++](./leetcode/652.find-duplicate-subtrees.cpp) | Other | \- | \- | - |
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| [Leetcode-366](https://leetcode.com/problems/find-leaves-of-binary-tree/) | Find Leaves Of Binary Tree | [c++](./leetcode/366.find-leaves-of-binary-tree.cpp), [python3](./leetcode/366.find-leaves-of-binary-tree.py) | Other | \- | \- | - |
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| [Leetcode-724](https://leetcode.com/problems/find-pivot-index/) | Find Pivot Index | [c++](./leetcode/724.find-pivot-index.cpp) | Other | \- | \- | - |
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| [Leetcode-277](https://leetcode.com/problems/find-the-celebrity/) | Find The Celebrity | [c++](./leetcode/277.find-the-celebrity.cpp), [python3](./leetcode/277.find-the-celebrity.py) | Other | \- | \- | - |
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| [Leetcode-41](https://leetcode.com/problems/first-missing-positive/) | First Missing Positive | [python3](./leetcode/41.first-missing-positive.py) | Other | \- | \- | - |
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| [Leetcode-387](https://leetcode.com/problems/first-unique-character-in-a-string/) | First Unique Character In A String | [c++](./leetcode/387.first-unique-character-in-a-string.cpp), [python3](./leetcode/387.first-unique-character-in-a-string.py) | Other | \- | \- | - |
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| [Leetcode-251](https://leetcode.com/problems/flatten-2d-vector/) | Flatten 2D Vector | [python3](./leetcode/251.flatten-2d-vector.py) | Other | \- | \- | - |

‎leetcode/277.find-the-celebrity.cpp

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‎leetcode/277.find-the-celebrity.py

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‎lintcode/645.find-the-celebrity.cpp

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// Tag: Simulation
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// Time: O(N)
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// Space: O(N)
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// Ref: Leetcode-277
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// Note: -
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// Suppose you are at a party with `n` people (labeled from `0` to `n - 1`) and among them, there may exist one celebrity.
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// The definition of a celebrity is that all the other `n - 1` people know him/her but he/she does not know any of them.
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//
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// Now you want to find out who the celebrity is or verify that there is not one.
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// The only thing you are allowed to do is to ask questions like: "Hi, A.
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// Do you know B?" to get information of whether A knows B.
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// You need to find out the celebrity (or verify there is not one) by asking as few questions as possible (in the asymptotic sense).
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//
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// You are given a helper function `bool knows(a, b)` which tells you whether A knows B.
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// Implement a function `int findCelebrity(n)`, your function should minimize the number of calls to `knows`.
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//
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// **Example1**
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// ```
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// Input:
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// 2 // next n * (n - 1) lines
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// 0 knows 1
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// 1 does not know 0
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// Output: 1
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// Explanation:
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// Everyone knows 1,and 1 knows no one.
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// ```
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// **Example2**
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// ```
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// Input:
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// 3 // next n * (n - 1) lines
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// 0 does not know 1
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// 0 does not know 2
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// 1 knows 0
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// 1 does not know 2
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// 2 knows 0
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// 2 knows 1
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// Output: 0
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// Explanation:
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// Everyone knows 0,and 0 knows no one.
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// 0 does not know 1,and 1 knows 0.
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// 2 knows everyone,but 1 does not know 2.
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// ```
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//
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// There will be exactly one celebrity if he/she is in the party. Return the celebrity's label if there is a celebrity in the party. If there is no celebrity, return `-1`.
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// Forward declaration of the knows API.
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bool knows(int a, int b);
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class Solution {
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public:
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/**
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* @param n a party with n people
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* @return the celebrity's label or -1
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*/
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int findCelebrity(int n) {
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// Write your code here
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stack <int> st;
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for (int i = 0; i < n; i++) {
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st.push(i);
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}
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while (st.size() > 1) {
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int a = st.top();
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st.pop();
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int b = st.top();
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st.pop();
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if (knows(a, b)) {
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st.push(b);
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} else {
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st.push(a);
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}
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}
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int celebrity = st.top();
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for (auto i = 0; i < n; ++i) {
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if (i == celebrity) {
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continue;
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}
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if(!knows(i, celebrity)) {
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return -1;
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}
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}
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return celebrity;
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}
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};
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class Solution {
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public:
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/**
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* @param n a party with n people
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* @return the celebrity's label or -1
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*/
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int findCelebrity(int n) {
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// Write your code here
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int candidate = 0;
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for(auto i = 1; i < n; ++i) {
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candidate = knows(candidate, i) ? i: candidate;
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}
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for (auto i = 0; i < n; ++i) {
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if (candidate == i) {
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continue;
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}
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if(knows(candidate, i)) {
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return -1;
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}
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if (!knows(i, candidate)) {
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return -1;
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}
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}
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return candidate;
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}
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};

‎lintcode/645.find-the-celebrity.py

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# Tag: Simulation
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# Time: O(N)
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# Space: O(N)
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# Ref: Leetcode-277
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# Note: -
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# Suppose you are at a party with `n` people (labeled from `0` to `n - 1`) and among them, there may exist one celebrity.
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# The definition of a celebrity is that all the other `n - 1` people know him/her but he/she does not know any of them.
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#
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# Now you want to find out who the celebrity is or verify that there is not one.
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# The only thing you are allowed to do is to ask questions like: "Hi, A.
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# Do you know B?" to get information of whether A knows B.
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# You need to find out the celebrity (or verify there is not one) by asking as few questions as possible (in the asymptotic sense).
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#
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# You are given a helper function `bool knows(a, b)` which tells you whether A knows B.
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# Implement a function `int findCelebrity(n)`, your function should minimize the number of calls to `knows`.
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#
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# **Example1**
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# ```
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# Input:
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# 2 // next n * (n - 1) lines
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# 0 knows 1
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# 1 does not know 0
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# Output: 1
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# Explanation:
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# Everyone knows 1,and 1 knows no one.
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# ```
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# **Example2**
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# ```
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# Input:
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# 3 // next n * (n - 1) lines
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# 0 does not know 1
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# 0 does not know 2
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# 1 knows 0
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# 1 does not know 2
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# 2 knows 0
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# 2 knows 1
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# Output: 0
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# Explanation:
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# Everyone knows 0,and 0 knows no one.
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# 0 does not know 1,and 1 knows 0.
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# 2 knows everyone,but 1 does not know 2.
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# ```
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#
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# There will be exactly one celebrity if he/she is in the party. Return the celebrity's label if there is a celebrity in the party. If there is no celebrity, return `-1`.
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"""
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The knows API is already defined for you.
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@param a, person a
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@param b, person b
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@return a boolean, whether a knows b
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you can call Celebrity.knows(a, b)
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"""
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class Solution:
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# @param {int} n a party with n people
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# @return {int} the celebrity's label or -1
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def findCelebrity(self, n):
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# Write your code here
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stack = [i for i in range(n)]
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while len(stack) > 1:
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a = stack.pop()
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b = stack.pop()
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if Celebrity.knows(a, b):
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stack.append(b)
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else:
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stack.append(a)
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celebrity = stack[0]
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for i in range(n):
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if i != celebrity and not Celebrity.knows(i, celebrity):
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return -1
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return celebrity
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class Solution:
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# @param {int} n a party with n people
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# @return {int} the celebrity's label or -1
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def findCelebrity(self, n):
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# Write your code here
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celebrity = 0
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for i in range(1, n):
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if Celebrity.knows(celebrity, i):
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celebrity = i
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for i in range(n):
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if i == celebrity:
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continue
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if Celebrity.knows(celebrity, i):
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return -1
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if not Celebrity.knows(i, celebrity):
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return -1
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return celebrity

‎list/.vscode/settings.json

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"string": "cpp",
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"string_view": "cpp",
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"unordered_map": "cpp",
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"vector": "cpp"
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"vector": "cpp",
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"queue": "cpp",
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"stack": "cpp"
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}
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}

‎list/9c-top.md

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17. https://leetcode.com/problems/longest-absolute-file-path/
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18. https://leetcode.com/problems/roman-to-integer/
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19. https://leetcode.com/problems/integer-to-roman/
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- https://www.lintcode.com/problem/identify-celebrity/
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20. https://www.lintcode.com/problem/identify-celebrity/
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### 基础、数据结构
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