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Update 0454.四数相加II.md

0454-四数相加,力扣函数参数更新,故更新对应题解代码

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Original file line number	Diff line number	Diff line change
`@@ -62,18 +62,18 @@ C++代码:`
`62`	`62`	```CPP
`63`	`63`	`class Solution {`
`64`	`64`	`public:`
`65`		`- int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {`
	`65`	`+ int fourSumCount(vector<int>& nums1, vector<int>& nums2, vector<int>& nums3, vector<int>& nums4) {`
`66`	`66`	`unordered_map<int, int> umap; //key:a+b的数值,value:a+b数值出现的次数`
`67`	`67`	`// 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`
`68`		`- for (int a : A) {`
`69`		`- for (int b : B) {`
	`68`	`+ for (int a : nums1) {`
	`69`	`+ for (int b : nums2) {`
`70`	`70`	`umap[a + b]++;`
`71`	`71`	`}`
`72`	`72`	`}`
`73`	`73`	`int count = 0; // 统计a+b+c+d = 0 出现的次数`
`74`	`74`	`// 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。`
`75`		`- for (int c : C) {`
`76`		`- for (int d : D) {`
	`75`	`+ for (int c : nums3) {`
	`76`	`+ for (int d : nums4) {`
`77`	`77`	`if (umap.find(0 - (c + d)) != umap.end()) {`
`78`	`78`	`count += umap[0 - (c + d)];`
`79`	`79`	`}`

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