454. 四数相加 II
题目:454. 四数相加 II
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 = 0
2. (1, 1, 0, 0) -> nums1[1] + nums2[1] + nums3[0] + nums4[0] = 2 + (-1) + (-1) + 0 = 0
示例 2:
输入:nums1 = [0], nums2 = [0], nums3 = [0], nums4 = [0]
输出:1
提示:
n == nums1.length
n == nums2.length
n == nums3.length
n == nums4.length
1 <= n <= 200
-228 <= nums1[i], nums2[i], nums3[i], nums4[i] <= 228
方法一:
看官方题解:https://leetcode.cn/problems/4sum-ii/solutions/499745/si-shu-xiang-jia-ii-by-leetcode-solution
go
func fourSumCount(nums1 []int, nums2 []int, nums3 []int, nums4 []int) int {
count := 0
m := make(map[int]int)
for _, n1 := range nums1 {
for _, n2 := range nums2 {
m[n1+n2]++
}
}
for _, n1 := range nums3 {
for _, n2 := range nums4 {
count += m[0-n1-n2]
}
}
return count
}
"官方也会写O(n²)的算法啊"