leetcode-454-4sum-ii
https://leetcode.com/problems/4sum-ii/description/
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[j]+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
给四个整型数组A.,B,C,D. 找出有多少对(i,j,k,l) 使得 (A[i]+B[j]+C[j]+D[l]) = 0. 假设四个数组的长度都一样。 四个数组。。怎么二分 
…. 三个加起来，分最后一个 N^3*logN…超时、
看题解： 将两个数组逐项求和。 保存和相同的个数到map中， 对另外两个数组逐项求和，与之前的和互为相反数的 即满足条件四个数组和为零. 从map中取出值累加就为最终结果 还有这种操作？ o(n^2)
class Solution {
public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
int num = 0;
int n = A.length;
Map<Integer,Integer> m = new HashMap<Integer,Integer>();
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
int sum = A[i]+B[j];
m.put(sum,m.getOrDefault(sum,0)+1);
}
}
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
int sum = -(C[i] + D[j]);
num += m.getOrDefault(sum,0);
}
}
return num;
}
}