add 213

Author: geemaple

README.md

@@ -6,7 +6,7 @@ My personal leetcode answers<br/>
 This is a **continually updated** open source project<br/>
 <br/>
 Total sovled: **343**<br/>
-Auto updated at: **2024-07-17 19:22:36**<br/>
+Auto updated at: **2024-07-19 10:58:47**<br/>
 
 ## 软件/Softwares
 - [Anki](https://apps.ankiweb.net/)
@@ -195,7 +195,7 @@ python problem.py https://www.lintcode.com/problem/92 -l cpp
 | [Leetcode-3048. Earliest Second To Mark Indices I](https://leetcode.com/problems/earliest-second-to-mark-indices-i/description/) | [python](./leetcode/3048.earliest-second-to-mark-indices-i.py), [c++](./leetcode/3048.earliest-second-to-mark-indices-i.cpp) | O\(M\*logM\) | O\(M\) | \- | - |
 
 ## Dynamic Programming
-| Problem(29) | Solution | Time | Space | Note | Ref |
+| Problem(30) | Solution | Time | Space | Note | Ref |
 | ----- | ----- | ----- | ----- | ----- | ----- |
 | [Leetcode-10. Regular Expression Matching](https://leetcode.com/problems/regular-expression-matching/description/) | [python](./leetcode/10.regular-expression-matching.py), [c++](./leetcode/10.regular-expression-matching.cpp) | O\(MN\) | O\(MN\) | \- | - |
 | [Leetcode-64. Minimum Path Sum](https://leetcode.com/problems/minimum-path-sum/description/) | [c++](./leetcode/64.minimum-path-sum.cpp), [python](./leetcode/64.minimum-path-sum.py) | O\(MN\) | O\(MN\) | \- | - |
@@ -208,6 +208,7 @@ python problem.py https://www.lintcode.com/problem/92 -l cpp
 | [Leetcode-139. Word Break](https://leetcode.com/problems/word-break/description/) | [c++](./leetcode/139.word-break.cpp), [python](./leetcode/139.word-break.py) | O\(MN\) | O\(N\) | \- | - |
 | [Leetcode-188. Best Time To Buy And Sell Stock IV](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/description/) | [python](./leetcode/188.best-time-to-buy-and-sell-stock-iv.py), [c++](./leetcode/188.best-time-to-buy-and-sell-stock-iv.cpp) | O\(NK\) | O\(NK\) | \- | - |
 | [Leetcode-198. House Robber](https://leetcode.com/problems/house-robber/description/) | [c++](./leetcode/198.house-robber.cpp), [python](./leetcode/198.house-robber.py) | O\(N\) | O\(N\) | \- | - |
+| [Leetcode-213. House Robber II](https://leetcode.com/problems/house-robber-ii/description/) | [c++](./leetcode/213.house-robber-ii.cpp), [python](./leetcode/213.house-robber-ii.py) | O\(N\) | O\(N\) | \- | - |
 | [Leetcode-221. Maximal Square](https://leetcode.com/problems/maximal-square/description/) | [python](./leetcode/221.maximal-square.py), [c++](./leetcode/221.maximal-square.cpp) | O\(MN\) | O\(MN\) | \- | - |
 | [Leetcode-279. Perfect Squares](https://leetcode.com/problems/perfect-squares/description/) | [c++](./leetcode/279.perfect-squares.cpp), [python](./leetcode/279.perfect-squares.py) | O\(n^\{3/2\}\) | O\(N\) | \- | - |
 | [Leetcode-300. Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence/description/) | [python](./leetcode/300.longest-increasing-subsequence.py), [c++](./leetcode/300.longest-increasing-subsequence.cpp) | O\(N\*logN\) | O\(N\) | LIS \| std::lower\_bound | - |
@@ -330,7 +331,7 @@ python problem.py https://www.lintcode.com/problem/92 -l cpp
 | [Lintcode-1790. Rotate String II](https://www.lintcode.com/problem/rotate-string-ii) | [c++](./lintcode/1790.rotate-string-ii.cpp), [python](./lintcode/1790.rotate-string-ii.py) | O\(N\) | O\(N\) | Simulation | - |
 
 ## Other
-| Problem(237) | Solution | Time | Space | Note | Ref |
+| Problem(236) | Solution | Time | Space | Note | Ref |
 | ----- | ----- | ----- | ----- | ----- | ----- |
 | [Leetcode-1. Two Sum](https://leetcode.com/problems/two-sum/description/) | [python](./leetcode/1.two-sum.py), [c++](./leetcode/1.two-sum.cpp) | \- | \- | \- | - |
 | [Leetcode-2. Add Two Numbers](https://leetcode.com/problems/add-two-numbers/description/) | [python](./leetcode/2.add-two-numbers.py), [c++](./leetcode/2.add-two-numbers.cpp) | \- | \- | \- | - |
@@ -429,7 +430,6 @@ python problem.py https://www.lintcode.com/problem/92 -l cpp
 | [Leetcode-210. Course Schedule II](https://leetcode.com/problems/course-schedule-ii/description/) | [c++](./leetcode/210.course-schedule-ii.cpp), [python](./leetcode/210.course-schedule-ii.py) | \- | \- | \- | - |
 | [Leetcode-211. Add And Search Word Data Structure Design](https://leetcode.com/problems/add-and-search-word-data-structure-design/description/) | [c++](./leetcode/211.add-and-search-word-data-structure-design.cpp), [python](./leetcode/211.add-and-search-word-data-structure-design.py) | \- | \- | \- | - |
 | [Leetcode-212. Word Search II](https://leetcode.com/problems/word-search-ii/description/) | [c++](./leetcode/212.word-search-ii.cpp), [python](./leetcode/212.word-search-ii.py) | \- | \- | \- | - |
-| [Leetcode-213. House Robber II](https://leetcode.com/problems/house-robber-ii/description/) | [c++](./leetcode/213.house-robber-ii.cpp), [python](./leetcode/213.house-robber-ii.py) | \- | \- | \- | - |
 | [Leetcode-217. Contains Duplicate](https://leetcode.com/problems/contains-duplicate/description/) | [c++](./leetcode/217.contains-duplicate.cpp) | \- | \- | \- | - |
 | [Leetcode-219. Contains Duplicate II](https://leetcode.com/problems/contains-duplicate-ii/description/) | [c++](./leetcode/219.contains-duplicate-ii.cpp) | \- | \- | \- | - |
 | [Leetcode-228. Summary Ranges](https://leetcode.com/problems/summary-ranges/description/) | [python](./leetcode/228.summary-ranges.py) | \- | \- | \- | - |

blog/leetcode-1.md

@@ -0,0 +1,96 @@
+---
+layout: post
+index: 1
+title: "LeetCode-1.两数之和(Two Sum)"
+categories: Leetcode
+tags: Leetcode Array HashTable
+excerpt: "起航，Great Voyage"
+---
+
+* content
+{:toc}
+
+相信每个想刷leetcode的人，第一次打开网站，就从这道题开始的。
+
+一起起航吧，Great Voyage.
+
+## 1. 两数之和：
+
+给定一个整数数组 nums 和一个目标值 target，请你在该数组中找出和为目标值的那 两个 整数，并返回他们的数组下标。
+
+你可以假设每种输入只会对应一个答案。但是，数组中同一个元素不能使用两遍。
+
+例如:
+```
+给定 nums = [2, 7, 11, 15], target = 9
+
+因为 nums[0] + nums[1] = 2 + 7 = 9
+所以返回 [0, 1]
+```
+
+来源：力扣（LeetCode）
+
+链接：[https://leetcode-cn.com/problems/two-sum/](https://leetcode-cn.com/problems/two-sum/)
+
+Link：[https://leetcode.com/problems/two-sum/](https://leetcode.com/problems/two-sum/)
+
+### 哈希
+
+O(N)
+
+遍历数组
+
+如果在字典找到数字，则输出[字典下标，当前下标]
+
+找不到答案，把当前的值放入字典(val -> index)
+
+因为题目说**有且只有一组答案**，所以不用判断数组为空，返回值不存在的情况
+
+```python
+class Solution:
+    def twoSum(self, nums: List[int], target: int) -> List[int]:
+        map = dict()
+        for i in range(len(nums)):
+            val = target - nums[i]
+            if val in map:
+                return [map[val], i]
+            else:
+                map[nums[i]] = i
+```
+
+### 双指针
+
+O(N*logN)
+
+如果题目要求返回两个具体数字，可以先排序，然后使用双指针(start, end)
+
+```res = nums[start] + nums[end]```
+
+如果**res < target**, 则说明**nums[start] + 最大的数**都小于target, start的值太小了，应该start++
+
+如果**res > target**, 则说明**nums[end] + 最小的数**都大于target, end的值太大了，应该end--
+
+如果**res = target**, 则输出答案
+
+```python
+class Solution(object):
+    def twoSum(self, nums: List[int], target: int) -> List[int]:
+        start = 0
+        end = len(nums) - 1
+        
+        nums.sort()
+        
+        while(start < end):
+            res = nums[start] + nums[end]
+            if res == target:
+                return (nums[start], nums[end])
+            elif res < target:
+                start += 1
+            else:
+                end -= 1
+
+```
+
+--End--
+
+

blog/leetcode-10.md

@@ -0,0 +1,203 @@
+---
+layout: post
+index: 10
+title: "LeetCode-10.正则表达式匹配(Regular Expression Matching)"
+categories: Leetcode
+tags: Leetcode Math DP Backtracking
+excerpt: "心态爆炸之旅"
+---
+
+* content
+{:toc}
+
+⚠️前方，高能预警
+
+这是第二道Hard的题目, 属于劝退类型
+
+## 10. 正则表达式匹配
+
+给你一个字符串 s 和一个字符规律 p，请你来实现一个支持 '.' 和 '*' 的正则表达式匹配。
+
+'.' 匹配任意单个字符
+
+'*' 匹配零个或多个前面的那一个元素
+
+所谓匹配，是要涵盖整个字符串s的，而不是部分字符串。
+
+说明:
+
+s 可能为空，且只包含从 a-z 的小写字母。
+
+p 可能为空，且只包含从 a-z 的小写字母，以及字符 . 和 *。
+
+示例 1:
+
+```
+输入:
+s = "aa"
+p = "a"
+输出: false
+解释: "a" 无法匹配 "aa" 整个字符串。
+```
+
+示例 2:
+
+```
+输入:
+s = "aa"
+p = "a*"
+输出: true
+解释: 因为 '*' 代表可以匹配零个或多个前面的那一个元素, 在这里前面的元素就是 'a'。因此，字符串 "aa" 可被视为 'a' 重复了一次。
+```
+
+示例 3:
+
+```
+输入:
+s = "ab"
+p = ".*"
+输出: true
+解释: ".*" 表示可匹配零个或多个（'*'）任意字符（'.'）。
+```
+
+示例 4:
+
+```
+输入:
+s = "aab"
+p = "c*a*b"
+输出: true
+解释: 因为 '*' 表示零个或多个，这里 'c' 为 0 个, 'a' 被重复一次。因此可以匹配字符串 "aab"。
+```
+
+示例 5:
+
+```
+输入:
+s = "mississippi"
+p = "mis*is*p*."
+输出: false
+```
+
+来源：力扣（LeetCode）
+
+链接：[https://leetcode-cn.com/problems/regular-expression-matching/](https://leetcode-cn.com/problems/regular-expression-matching/)
+
+Link：[https://leetcode.com/problems/regular-expression-matching/](https://leetcode.com/problems/regular-expression-matching/)
+
+### 暴力破解
+
+将字符串p中的字符*全排列展开，直到p与s长度相等为止，判断s是否和p一致
+
+判断过程中，由于'.'可以匹配任意字符串，可以认为相等
+
+### 动态规划
+
+O(N^2)
+
+定义xxx代表前i个字符串已经能够匹配，这里需要两个变量(i, j)分别代表字符串s和p
+
+table[i][j] = True, 代表s前i个字符能够和p前j个字符匹配
+
+#### 思考步骤
+
+动态规划，先从最后一步倒推，假设字符串s和p能够匹配，那么最后一步可能是
+
+> 下图中认为xxxxxxx部分已完成判断
+
+条件1, 最后一步, s和p最后两个字母相等**或者**p最后一个字母是'.' 
+
+```python
+xxxxxxxxxxxx  a
+xxxxxxxxxxxx  a
+
+xxxxxxxxxxxx  a
+xxxxxxxxxxxx  .
+
+table[i][j] = ((p[i - i] == s[i - i] or p[i - i] == '.') and table[i - 1][j - 1])
+```
+
+条件2, 最后一步, p最后一个是\*, 无论是否能够匹配, 字符\*都可以零次展开
+
+```python
+xxxxxxxxxxx x  a
+xxxxxxxxxxx a  *
+            ^
+
+xxxxxxxxxxx x  a
+xxxxxxxxxxx c  *
+            ^
+# 0次展开，消耗p字符串*前面的一个字母, 用^来标记, 但s保持不变
+
+table[i][j] = table[i][j - 2]
+```
+
+条件3, p最后一个是\*, 当\*一次展开时: (0和1次展开, 就可以覆盖0...k次的情况)
+
+```python                                      
+xxxxxxxxxxx x  a          xxxxxxxxxxx x  a
+                                         ^
+xxxxxxxxxxx a  *          xxxxxxxxxxx a  a  *
+                                         
+                    ->
+xxxxxxxxxxx x  a          xxxxxxxxxxx x  a
+                                         ^
+xxxxxxxxxxx .  *          xxxxxxxxxxx .  .  *       
+
+# 1次展开，消耗掉s字符串^对应的字母，但是p保持不变
+
+table[i][j] = (p[j - 2] == s[i] or p[i - 2] == '.') and table[i - 1][j])
+```
+
+所有条件，只要有一个可以匹配，就可以认为匹配
+
+#### 计算方向
+
+从左到右，从比较短的字符开始，这样长字符，能够利用上短字符的计算结果
+
+#### 边界条件
+
+初始化table内容为False, 矩阵行列记得多开一个空间
+
+其中table[0][0] = True, 代表两个空字符串可以匹配
+
+table[i][0] = False, 也就是p为空，那除了s也为空外，都不能匹配
+
+> ⚠️注意，前i个字符，对应的当前下标是i - 1。
+
+```python
+
+class Solution:
+    def isMatch(self, s: str, p: str) -> bool:
+        m = len(s)
+        n = len(p)
+        table = [[False for _ in range(n + 1)] for _ in range(m + 1)]
+
+        for i in range(m + 1):
+            for j in range(n + 1):
+                if i == 0 and j == 0:
+                    table[i][j] = True
+                    continue
+
+                if j == 0:
+                    table[i][j] = False
+                    continue
+
+                if p[j - 1] != '*':
+                    if i - 1 >= 0 and (p[j - 1] == '.' or p[j - 1] == s[i- 1]):
+                        table[i][j] = table[i - 1][j - 1] # 尾字母可匹配
+                else:
+                    if j - 2 >= 0:
+                        table[i][j] = table[i][j - 2] # 0次展开
+
+                        if i - 1 >= 0 and (p[j - 2] == '.' or p[j - 2] == s[i - 1]):
+                            table[i][j] = table[i][j] or table[i - 1][j] # 1次展开
+
+        return table[m][n]
+```
+
+--End--
+
+难度爆炸，随便的草稿，不值得细看
+
+![草稿说明]({{site.static}}/images/leetcode-sketch-algorithm-10.png)