标签:tuples 4sum ati lis contain inpu com lang hashmap
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
给定四个数组,在各个数组中取出一个数使其和为0,求这样的数对的个数。
用HashMap记录其中两个数组的组合情况,再遍历另外两个数组的组合进行处理。
class Solution {
public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
int count = 0;
Map<Integer, Integer> hash = new HashMap<>();
for (int i = 0; i < A.length; i++) {
for (int j = 0; j < B.length; j++) {
int sum = A[i] + B[j];
hash.put(sum, hash.getOrDefault(sum, 0) + 1);
}
}
for (int i = 0; i < C.length; i++) {
for (int j = 0; j < D.length; j++) {
int target = - C[i] - D[j];
if (hash.containsKey(target)) {
count += hash.get(target);
}
}
}
return count;
}
}
标签:tuples 4sum ati lis contain inpu com lang hashmap
原文地址:https://www.cnblogs.com/mapoos/p/14150426.html