2019/10/25

Author: rchen102

Python/leetcode 1.py

@@ -5,5 +5,4 @@ def twoSum(self, nums, target):
             if (target - nums[i]) in dic:
                 return [dic[target - nums[i]], i]
             else:
-                dic[nums[i]] = i
-        
+                dic[nums[i]] = i

Python/leetcode 15.py

@@ -0,0 +1,31 @@
+class Solution:
+    def threeSum(self, nums: List[int]) -> List[List[int]]:
+        res = []
+        nums.sort()
+        for i in range(len(nums)-2):
+            if i == 0 or (i > 0 and nums[i] != nums[i-1]):
+                self.twoSum(res, i, nums)
+        return res    
+    
+    def twoSum(self, res, idx, nums):
+        target = -nums[idx]
+        lo, hi = idx + 1, len(nums) - 1
+        while (lo < hi):
+            if nums[lo] + nums[hi] == target:
+                res.append((nums[idx], nums[lo], nums[hi]))
+                while (lo < hi) and (nums[lo] == nums[lo+1]):
+                    lo += 1
+                while (lo < hi) and (nums[hi] == nums[hi-1]):
+                    hi -= 1
+                lo += 1
+                hi -= 1
+            elif nums[lo] + nums[hi] < target:
+                lo += 1
+            else:
+                hi -= 1
+    
+
+
+    
+        
+        

总结/Array/Array.MD renamed to SUMMARY/Array/Array.MD

@@ -0,0 +1,9 @@
+## 1 Two Sum
+
+### Solution1: Brute force
+T: O(n^2) S: O(1)
+
+双重循环查找
+
+### Solution2: HashMap
+T: O(n) S: O(n)

SUMMARY/Array/Leetcode 1.MD

@@ -0,0 +1,8 @@
+## 15 3Sum
+
+### Solution1: Sort + Two pointer
+T： O(n^2) S: O(1)
+
+- 避免重复，首先进行排序
+- 利用Two Pointers实现增强版Two Sum
+    + 不使用HashMap的原因: 节省时间+避免重复

SUMMARY/Array/Leetcode 15.MD

@@ -0,0 +1 @@
+1. 输入数据中含有重复数字，输出要求含有避免重复 -> 排序