[LeetCode] 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:2Explanation:The two tuples are:1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 02. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0

分成两半用哈希，这样时间复杂度正好是o(n**2)，分成3和1会超时

超时代码：

public class Solution2 {public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {  HashMap<Integer,Integer> map=new HashMap<Integer,Integer>();  for(int a:A){  for(int b:B){  for(int c:C){  int sum=a+b+c;  Integer num=map.get(sum);  if(num==null) num=0;  map.put(sum, num+1);  }  }  }  int re=0;  for(int d:D){  if(map.containsKey(-d)) re+=map.get(-d);  }  return re;  }}

ac代码

public class Solution {  public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {  HashMap<Integer,Integer> map=new HashMap<Integer,Integer>();  for(int a:A){  for(int b:B){  int sum=a+b;  Integer num=map.get(sum);  if(num==null) num=0;  map.put(sum, num+1);  }  }  int re=0;  for(int c:C){  for(int d:D){  int sum=c+d;  if(map.containsKey(-sum)) re+=map.get(-sum);  }  }  return re;  }}