# 18. 四数之和 (opens new window)
# Description
Difficulty: 中等
Related Topics: 数组 (opens new window), 双指针 (opens new window), 排序 (opens new window)
给你一个由 n
个整数组成的数组 nums
,和一个目标值 target
。请你找出并返回满足下述全部条件且不重复的四元组 [nums[a], nums[b], nums[c], nums[d]]
(若两个四元组元素一一对应,则认为两个四元组重复):
0 <= a, b, c, d < n
a
、b
、c
和d
互不相同nums[a] + nums[b] + nums[c] + nums[d] == target
你可以按 任意顺序 返回答案 。
示例 1:
输入:nums = [1,0,-1,0,-2,2], target = 0
输出:[[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]
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2
示例 2:
输入:nums = [2,2,2,2,2], target = 8
输出:[[2,2,2,2]]
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提示:
1 <= nums.length <= 200
- -109 <= nums[i] <= 109
- -109 <= target <= 109
# Solution
Language: JavaScript
/**
* @param {number[]} nums
* @param {number} target
* @return {number[][]}
*/
var fourSum = function(nums, target) {
nums.sort((a, b) => a - b)
const result = []
const n = nums.length
for (let i = 0;i < n - 3;i++) {
if (i != 0 && nums[i] === nums[i - 1]) continue
for (let j = i + 1;j < n - 2;j++) {
if (j > i + 1 && nums[j] === nums[j - 1]) continue
let p = j + 1
let q = n - 1
while (p < q) {
if (p > j + 1 && nums[p] === nums[p - 1]) {
p++
continue
}
if (q < n - 1 && nums[q] === nums[q + 1]) {
q--
continue
}
const sum = nums[i] + nums[j] + nums[p] + nums[q]
if (sum === target) {
result.push([nums[i], nums[j], nums[p], nums[q]])
p++
q--
} else if (sum < target) {
p++
} else {
q--
}
}
}
}
return result
};
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# 通用解法
var fourSum = function(nums, target) {
nums.sort((a, b) => a - b)
const len = nums.length
return nSum(nums, 4, 0, target, len)
function nSum(nums, n, start, target, len) {
if (len < n || n < 2) return []
const result = []
if (n === 2) {
let i = start
let j = len - 1
while (i < j) {
if (i > start && nums[i] === nums[i - 1]) {
i++
continue
}
if (j < len - 1 && nums[j] === nums[j + 1]) {
j--
continue
}
const sum = nums[i] + nums[j]
if (target === sum) {
result.push([nums[i], nums[j]])
i++
j--
} else if (sum < target) {
i++
} else {
j--
}
}
} else {
for(let i = start;i < len;i++) {
if (i > start && nums[i] === nums[i - 1]) continue
const resultArr = nSum(nums, n - 1, i + 1, target - nums[i], len)
for (let arr of resultArr) {
result.push([nums[i], ...arr])
}
}
}
return result
}
}
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