20211108-04 Four Sum

https://leetcode-cn.com/problems/4sum/submissions/

思路：n个数之和，n-2重for循环 + 双指针

python

class Solution(object):
    def fourSum(self, nums, target):
        if len(nums) < 4:
            return []
        
        res = []
        nums.sort()
        for i in range(len(nums) - 3): # 第一个元素只需要遍历 0...n-3 就行
            if i  > 0 and nums[i] == nums[i-1]:
                continue
            
            for j in range(i+1, len(nums) - 2):
                if j > i + 1 and nums[j] == nums[j-1]:
                    continue
                
                new_tgt = target - nums[i] - nums[j]
                left, right = j + 1, len(nums) - 1
                while left < right:
                    if nums[left] + nums[right] == new_tgt:
                        res.append( [nums[i], nums[j], nums[left], nums[right]] )

                        left, right = left + 1, right - 1
                        while left < right and nums[left] == nums[left - 1]:
                            left += 1
                        
                        while left < right and nums[right] == nums[right + 1]:
                            right -= 1
                        
                    elif nums[left] + nums[right] < new_tgt:
                        left += 1
                    else:
                        right -= 1
        
        return res

标签：right,tgt,04,nums,res,Sum,len,Four,left
来源： https://www.cnblogs.com/dundundun/p/15522993.html