本文主要是介绍454. 4Sum II leetcode binary search,希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!
(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:
Input: A = [ 1, 2] B = [-2,-1] C = [-1, 2] D = [ 0, 2]Output: 2Explanation: 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
这一题的解法首先用一个map存任意两个数组的和的组合,然后在循环遍历另外两个数组的时候,从map中判断相关的数是否存在。
int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {//将四数转变为两个部分,首先遍历AB的组合(任意两个都可以),存下他们组合后的和的情况,然后遍历CD(另外两个)的和,看之前AB遍历的组合里有没有与此时值相反的值,有的话就加上AB中这个相反数出现的次数。//不过该题应该四个数组的长度都一样,所以前面以下五句可以简化为/*int len = A.size();if(0 == len){return 0;}*/int lenA = A.size();int lenB = B.size();int lenC = C.size();int lenD = D.size();if(0 == lenA || 0 == lenB || 0 == lenC || 0 == lenD) return 0;//以下是将A和B的所有的元素可能的和的数对放入一个map中去,注意leetcode的编辑器不包括hash_map//注意:如果A=[2,2],B=[-1.-1],那么ab的map的键只有1的情况,值为:2*2=4map<int,int> ab;for(int i = 0; i < lenA; i++)for(int j = 0; j < lenB; j++){if(ab.find(A[i]+B[j]) == ab.end())ab[A[i]+B[j]] = 1;elseab[A[i]+B[j]]++;}//以下是遍历CD然后去查找看看是否ab中含有-CD的项目int result = 0;for(int i = 0; i < lenC; i++)for(int j = 0; j < lenD; j++){int tmp = -(C[i]+D[j]);if(ab.find(tmp) != ab.end()){result += ab[tmp];}}return result;}
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