添加 problem 454

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
+}

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")
+}

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。
+
+
+