题目描述

详见3Sum思路
(排序，双指针)

排序后固定两个数，然后用双指针(left, right)去找3项, 4项

最后nums[i] + nums[j] + 3项 + 4项 = target

Python代码

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):
            # 防止重复 数组进入 res
            if i > 0 and nums[i] == nums[i-1]:
                continue
            # 当数组最小值和都大于target 跳出
            if nums[i] + nums[i+1] + nums[i+2] + nums[i+3] > target:
                break
            # 当数组最大值和都小于target,说明i这个数还是太小,遍历下一个
            if nums[i] + nums[n-1] + nums[n-2] + nums[n-3] < target:
                continue
            for j in range(i+1,n-2):
                # 防止重复 数组进入 res
                if j - i > 1 and nums[j] == nums[j-1]:
                    continue
                # 同理
                if nums[i] + nums[j] + nums[j+1] + nums[j+2] > target:
                    break
                # 同理
                if nums[i] + nums[j] + nums[n-1] + nums[n-2] < target:
                    continue
                # 双指针
                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

c++ y总版 【三重循环，本质是一样，过程写法区别】

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        sort(nums.begin(), nums.end());
        vector<vector<int>> res;
        for (int i = 0; i < nums.size(); i ++ ) {
            if (i && nums[i] == nums[i - 1]) continue;
            for (int j = i + 1; j < nums.size(); j ++ ) {
                if (j > i + 1 && nums[j] == nums[j - 1]) continue;
                for (int k = j + 1, u = nums.size() - 1; k < u; k ++ ) {
                    if (k > j + 1 && nums[k] == nums[k - 1]) continue;
                    while (u - 1 > k && nums[i] + nums[j] + nums[k] + nums[u - 1] >= target) u -- ;
                    if (nums[i] + nums[j] + nums[k] + nums[u] == target) {
                        res.push_back({nums[i], nums[j], nums[k], nums[u]});
                    }
                }
            }
        }

        return res;
    }
};