# leetcode #18 4Sum

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note: The solution set must not contain duplicate quadruplets.

``````For example, given array S = [1, 0, -1, 0, -2, 2], and target = 0.

A solution set is:
[
[-1,  0, 0, 1],
[-2, -1, 1, 2],
[-2,  0, 0, 2]
]
``````
• 题目大意
给定一组整数，找到数组中所有相加等于target的4个数，结果中不能有相同的一组。
与之前的3Sum 算法完全一样，只是多加了一层循环，时间复杂度由O(n2) 变为 O(n3)

``````/**
* @param {number[]} nums
* @param {number} target
* @return {number[][]}
*/
const fourSum = function (nums, target) {
nums.sort((a, b) => a - b);
const result = [];
for (let i = 0; i < nums.length - 3; i++) {
if (nums[i] + nums[i + 1] + nums[i + 2] + nums[i + 3] > target) break;  //可能得到的最小值比target还要大
if (nums[i] + nums[nums.length - 1] + nums[nums.length - 2] + nums[nums.length - 3] < target) continue;  //可能得到的最大值比target还要小
if (i > 0 && nums[i] === nums[i - 1]) { // 跳过相同的值
continue;
}
for (let l = i + 1; l < nums.length - 2; l++) {
if (nums[i] + nums[l + 1] + nums[l + 2] + nums[l] > target) break;
if (nums[i] + nums[nums.length - 1] + nums[nums.length - 2] + nums[l] < target) continue;
if (l > i + 1 && nums[l] === nums[l - 1]) { // 跳过相同的值
continue;
}
let j = l + 1;
let k = nums.length - 1;
const rest = target - nums[i] - nums[l];
while (j < k) {
if (nums[j] + nums[k] === rest) {
result.push([nums[i], nums[l], nums[j], nums[k]]);
while (j < k && nums[j] === nums[j + 1]) j++; // 跳过相同的值
while (j < k && nums[k] === nums[k - 1]) k--;
j++;
k--;
} else if (nums[j] + nums[k] > rest) k--; // 如果两数相加比剩余的值大，减小较大的值
else {
j++; // 否则增加较小的值
}
}
}
}
return result;
};

``````