leetcode题目 https://leetcode.com/problems/4sum-ii/
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:
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答案解析
public static int fourSumCount(int[] A, int[] B, int[] C, int[] D) { Map<Integer,Integer> map=new HashMap<>(16); for(int i=0;i<A.length;i++){ for(int j=0;j<B.length;j++){ int sum=A[i]+B[j]; Integer o = map.get(sum); if(o==null){ map.put(sum,1); }else{ map.put(sum,o+1); } } } /** * 建立hashmap 统计前两个数组累加和出现的次数 */ System.out.println(map); int res=0; for(int p=0;p<C.length;p++){ for(int q=0;q<D.length;q++){ int sum2=C[p]+D[q]; Integer integer = map.get(-sum2); if(integer==null){ res+=0; }else{ res+=integer; } } } return res;参考资料:
https://blog.csdn.net/qq508618087/article/details/53255924
https://blog.csdn.net/crystal_ting/article/details/76596188
