198 - 18 四数之和

题目

给定一个包含 n 个整数的数组 nums 和一个目标值 target，判断 nums 中是否存在四个元素 a，b，c 和 d ，使得 a + b + c + d 的值与 target 相等？找出所有满足条件且不重复的四元组。

注意：

答案中不可以包含重复的四元组。

示例：

给定数组 nums = [1, 0, -1, 0, -2, 2]，和 target = 0。

满足要求的四元组集合为： [ [-1, 0, 0, 1], [-2, -1, 1, 2], [-2, 0, 0, 2] ]

解答

排序，然后双指针，两层循环

递归

https://leetcode.com/problems/4sum/discuss/8545/Python-140ms-beats-100-and-works-for-N-sum-(Ngreater2)

class Solution:
    def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
        nums.sort()
        results = []
        def findNsum(l, r, target, n, result, results):
            if r-l+1<n or n<2 or target < nums[l]*n or target > nums[r]*n:
                return
            if n ==2:
                while l < r:
                    s = nums[l]+nums[r]
                    if s == target:
                        results.append(result+[nums[l], nums[r]])
                        l+=1
                        while l<r and nums[l]==nums[l-1]:
                            l+=1
                    elif s < target:
                        l+=1
                    else:
                         r-=1
            else:
                for i in range(l, r+1):
                    if i == l or (i > l and nums[i-1] != nums[i]):
                        findNsum(i+1, r, target-nums[i], n-1,
                                result+[nums[i]],results)
        findNsum(0, len(nums)-1, target, 4, [], results)
        return res

Runtime: 84 ms, faster than 92.68% of Python3 online submissions for 4Sum.

Memory Usage: 12.8 MB, less than 100.00% of Python3 online submissions for 4Sum.

双指针

https://leetcode-cn.com/problems/4sum/solution/gu-ding-liang-ge-shu-yong-shuang-zhi-zhen-zhao-lin/

class Solution:
    def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
        n = len(nums)
        if n < 4:
            return []
        nums.sort()
        res = []
        for i in range(n-3):
            if i > 0 and nums[i] == nums[i-1]:
                continue
            if nums[i] + nums[i+1] + nums[i+2] + nums[i+3] > target:
                break
            if nums[i] + nums[n-1] + nums[n-2] + nums[n-3] < target:
                continue
            for j in range(i+1, n-2):
                if j - i > 1 and nums[j] == nums[j-1]:
                    continue
                if nums[i] + nums[j] + nums[j+1] + nums[j+2] > target:
                    break
                if nums[i] + nums[j] + nums[n-1] + nums[n-2] < target:
                    continue
                left = j + 1
                right = n - 1
                while left < right:
                    tmp = nums[i] + nums[j] + nums[left] + nums[right]
                    if tmp == target:
                        res.append([nums[i], nums[j], nums[left], nums[right]])
                        while left < right and nums[left] == nums[left+1]:
                            left += 1
                        while left < right and nums[right] == nums[right-1]:
                            right -= 1
                        left += 1
                        right -= 1
                    elif tmp > target:
                        right -= 1
                    else:
                        left += 1
        return res

Runtime: 92 ms, faster than 86.42% of Python3 online submissions for 4Sum.

Memory Usage: 12.7 MB, less than 100.00% of Python3 online submissions for 4Sum.