diff --git a/Algorithms/0454. 4Sum II/454. 4Sum II.go b/Algorithms/0454. 4Sum II/454. 4Sum II.go new file mode 100644 index 00000000..9ec5e153 --- /dev/null +++ b/Algorithms/0454. 4Sum II/454. 4Sum II.go @@ -0,0 +1,18 @@ +package leetcode + +func fourSumCount(A []int, B []int, C []int, D []int) int { + m := make(map[int]int, len(A)*len(B)) + for _, a := range A { + for _, b := range B { + m[a+b]++ + } + } + ret := 0 + for _, c := range C { + for _, d := range D { + ret += m[0-c-d] + } + } + + return ret +} diff --git a/Algorithms/0454. 4Sum II/454. 4Sum II_test.go b/Algorithms/0454. 4Sum II/454. 4Sum II_test.go new file mode 100644 index 00000000..54fc5b42 --- /dev/null +++ b/Algorithms/0454. 4Sum II/454. 4Sum II_test.go @@ -0,0 +1,45 @@ +package leetcode + +import ( + "fmt" + "testing" +) + +type question454 struct { + para454 + ans454 +} + +// para 是参数 +// one 代表第一个参数 +type para454 struct { + a []int + b []int + c []int + d []int +} + +// ans 是答案 +// one 代表第一个答案 +type ans454 struct { + one int +} + +func Test_Problem454(t *testing.T) { + + qs := []question454{ + + question454{ + para454{[]int{1, 2}, []int{-2, -1}, []int{-1, 2}, []int{0, 2}}, + ans454{2}, + }, + } + + fmt.Printf("------------------------Leetcode Problem 454------------------------\n") + + for _, q := range qs { + _, p := q.ans454, q.para454 + fmt.Printf("【input】:%v 【output】:%v\n", p, fourSumCount(p.a, p.b, p.c, p.d)) + } + fmt.Printf("\n\n\n") +} diff --git a/Algorithms/0454. 4Sum II/README.md b/Algorithms/0454. 4Sum II/README.md new file mode 100644 index 00000000..84956bf0 --- /dev/null +++ b/Algorithms/0454. 4Sum II/README.md @@ -0,0 +1,37 @@ +# [454. 4Sum II](https://leetcode.com/problems/4sum-ii/) + +## 题目 + +Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero. + +To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range of -228 to 228 - 1 and the result is guaranteed to be at most 231 - 1. + +Example 1: + +```c +Input: +A = [ 1, 2] +B = [-2,-1] +C = [-1, 2] +D = [ 0, 2] + +Output: +2 + +Explanation: +The two tuples are: +1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0 +2. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0 +``` + + +## 题目大意 + +给出 4 个数组,计算这些数组中存在几对 i,j,k,l 可以使得 A[i] + B[j] + C[k] + D[l] = 0 。 + +## 解题思路 + +这道题的数据量不大,0 ≤ N ≤ 500,但是如果使用暴力解法,四层循环,会超时。这道题的思路和第 1 题思路也类似,先可以将 2 个数组中的组合都存入 map 中。之后将剩下的 2 个数组进行 for 循环,找出和为 0 的组合。这样时间复杂度是 O(n^2)。当然也可以讲剩下的 2 个数组的组合也存入 map 中,不过最后在 2 个 map 中查找结果也是 O(n^2) 的时间复杂度。 + + +