给你四个整数数组 nums1、nums2、nums3 和 nums4 ,数组长度都是 n ,请你计算有多少个元组 (i, j, k, l) 能满足:
- 0 <= i, j, k, l < n
- nums1[i] + nums2[j] + nums3[k] + nums4[l] == 0
示例1:
输入:nums1 = [1,2], nums2 = [-2,-1], nums3 = [-1,2], nums4 = [0,2]输出:2解释:两个元组如下:1. (0, 0, 0, 1) -> nums1[0] + nums2[0] + nums3[0] + nums4[1] = 1 + (-2) + (-1) + 2 = 02. (1, 1, 0, 0) -> nums1[1] + nums2[1] + nums3[0] + nums4[0] = 2 + (-1) + (-1) + 0 = 0
解析:判断前两个数组和跟后两个数组和取反
const a =[1,2]const b =[-1,-5]const c =[3,5]const d =[-3,-2]const valueMap = new Map()let count = 0a.forEach(i => {b.forEach(j => {valueMap.set(i+j,(valueMap.get(i+j) ||0 ) + 1)})})c.forEach(i => {d.forEach(j => {count+= (valueMap.get(0-i-j))||0})})console.log(count)
