Update 0015.三数之和.md

problems/0015.三数之和.md

@@ -298,61 +298,72 @@ class Solution {
 ```
 
 Python：
+（版本一） 双指针
 ```Python
 class Solution:
-    def threeSum(self, nums):
-        ans = []
-        n = len(nums)
+    def threeSum(self, nums: List[int]) -> List[List[int]]:
+        result = []
         nums.sort()
-	# 找出a + b + c = 0
-        # a = nums[i], b = nums[left], c = nums[right]
-        for i in range(n):
-            left = i + 1
-            right = n - 1
-	    # 排序之后如果第一个元素已经大于零，那么无论如何组合都不可能凑成三元组，直接返回结果就可以了
-            if nums[i] > 0: 
-                break
-            if i >= 1 and nums[i] == nums[i - 1]: # 去重a
+        
+        for i in range(len(nums)):
+            # 如果第一个元素已经大于0，不需要进一步检查
+            if nums[i] > 0:
+                return result
+            
+            # 跳过相同的元素以避免重复
+            if i > 0 and nums[i] == nums[i - 1]:
                 continue
-            while left < right:
-                total = nums[i] + nums[left] + nums[right]
-                if total > 0:
-                    right -= 1
-                elif total < 0:
+                
+            left = i + 1
+            right = len(nums) - 1
+            
+            while right > left:
+                sum_ = nums[i] + nums[left] + nums[right]
+                
+                if sum_ < 0:
                     left += 1
+                elif sum_ > 0:
+                    right -= 1
                 else:
-                    ans.append([nums[i], nums[left], nums[right]])
-		    # 去重逻辑应该放在找到一个三元组之后，对b 和 c去重
-                    while left != right and nums[left] == nums[left + 1]: left += 1
-                    while left != right and nums[right] == nums[right - 1]: right -= 1
-                    left += 1
+                    result.append([nums[i], nums[left], nums[right]])
+                    
+                    # 跳过相同的元素以避免重复
+                    while right > left and nums[right] == nums[right - 1]:
+                        right -= 1
+                    while right > left and nums[left] == nums[left + 1]:
+                        left += 1
+                        
                     right -= 1
-        return ans
+                    left += 1
+                    
+        return result
 ```
-Python (v3):
+（版本二） 使用字典
 
 ```python
 class Solution:
     def threeSum(self, nums: List[int]) -> List[List[int]]:
-        if len(nums) < 3: return []
-        nums, res = sorted(nums), []
-        for i in range(len(nums) - 2):
-            cur, l, r = nums[i], i + 1, len(nums) - 1
-            if res != [] and res[-1][0] == cur: continue # Drop duplicates for the first time.
-
-            while l < r:
-                if cur + nums[l] + nums[r] == 0:
-                    res.append([cur, nums[l], nums[r]])
-                    # Drop duplicates for the second time in interation of l & r. Only used when target situation occurs, because that is the reason for dropping duplicates.
-                    while l < r - 1 and nums[l] == nums[l + 1]:
-                        l += 1
-                    while r > l + 1 and nums[r] == nums[r - 1]:
-                        r -= 1
-                if cur + nums[l] + nums[r] > 0:
-                    r -= 1
+        result = []
+        nums.sort()
+        # 找出a + b + c = 0
+        # a = nums[i], b = nums[j], c = -(a + b)
+        for i in range(len(nums)):
+            # 排序之后如果第一个元素已经大于零，那么不可能凑成三元组
+            if nums[i] > 0:
+                break
+            if i > 0 and nums[i] == nums[i - 1]: #三元组元素a去重
+                continue
+            d = {}
+            for j in range(i + 1, len(nums)):
+                if j > i + 2 and nums[j] == nums[j-1] == nums[j-2]: # 三元组元素b去重
+                    continue
+                c = 0 - (nums[i] + nums[j])
+                if c in d:
+                    result.append([nums[i], nums[j], c])
+                    d.pop(c) # 三元组元素c去重
                 else:
-                    l += 1
-        return res
+                    d[nums[j]] = j
+        return result
 ```
 
 Go：