Commit 784437b

authored

Merge pull request #2907 from zenwangzy/master

Update 0454.四数相加II,力扣函数参数更新,故更新对应题解代码

2 parents a6caded + d6e6804 commit 784437bCopy full SHA for 784437b

File tree

-5

lines changed

-5

lines changed

Lines changed: 5 additions & 5 deletions

Original file line number	Diff line number	Diff line change
`@@ -60,18 +60,18 @@ C++代码:`
`60`	`60`	```CPP
`61`	`61`	`class Solution {`
`62`	`62`	`public:`
`63`		`- int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {`
	`63`	`+ int fourSumCount(vector<int>& nums1, vector<int>& nums2, vector<int>& nums3, vector<int>& nums4) {`
`64`	`64`	`unordered_map<int, int> umap; //key:a+b的数值,value:a+b数值出现的次数`
`65`	`65`	`// 遍历大A和大B数组,统计两个数组元素之和,和出现的次数,放到map中`
`66`		`- for (int a : A) {`
`67`		`- for (int b : B) {`
	`66`	`+ for (int a : nums1) {`
	`67`	`+ for (int b : nums2) {`
`68`	`68`	`umap[a + b]++;`
`69`	`69`	`}`
`70`	`70`	`}`
`71`	`71`	`int count = 0; // 统计a+b+c+d = 0 出现的次数`
`72`	`72`	`// 再遍历大C和大D数组,找到如果 0-(c+d) 在map中出现过的话,就把map中key对应的value也就是出现次数统计出来。`
`73`		`- for (int c : C) {`
`74`		`- for (int d : D) {`
	`73`	`+ for (int c : nums3) {`
	`74`	`+ for (int d : nums4) {`
`75`	`75`	`if (umap.find(0 - (c + d)) != umap.end()) {`
`76`	`76`	`count += umap[0 - (c + d)];`
`77`	`77`	`}`

Comments

(0)