454. 4Sum II
作者:互联网
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
class Solution { public int fourSumCount(int[] A, int[] B, int[] C, int[] D) { int res = 0; int le = A.length; if(le == 0) return res; Map<Integer, Integer> map = new HashMap(); for(int i = 0; i < le; i++){ for(int j = 0; j < le; j++){ int t = - A[i] - B[j]; map.put(t, map.getOrDefault(t, 0) + 1); } } for(int i = 0; i < le; i++){ for(int j = 0; j < le; j++){ int t = C[i] + D[j]; if(map.containsKey(t)) res += map.get(t); } } return res; } }
比4sum简单,因为不用考虑到重复元素(元素重复但是下标不重复)
解法很巧妙,用hashmap存放前两个数组每个pair和出现的次数,
然后在第三第四个数组中找有无一样的值,如果有就加上该次数,因为次数是AB数组独特的所以可以直接加。
标签:4Sum,le,map,int,res,454,II,++,数组 来源: https://www.cnblogs.com/wentiliangkaihua/p/11847925.html