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 -2^28 to 2^28 - 1 and the result is guaranteed to be at most 2^31 - 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
析：可以每两个数求和，以空间换时间
public int fourSumCount(int[] A, int[] B, int[] C, int[] D) { int res=0; Map<Integer, Integer> map1 = new HashMap<>(); Map<Integer, Integer> map2 = new HashMap<>(); for(int i=0;i<A.length;i++) for(int j=0;j<B.length;j++){ int sum=A[i]+B[j]; if(!map1.containsKey(sum)) map1.put(sum, 1); else map1.replace(sum, map1.get(sum), map1.get(sum)+1); } for(int i=0;i<C.length;i++) for(int j=0;j<D.length;j++){ int sum=C[i]+D[j]; if(!map2.containsKey(sum)) map2.put(sum, 1); else map2.replace(sum, map2.get(sum), map2.get(sum)+1); } for(int t:map1.keySet()) if(map1.containsKey(-t)) res+=map1.get(t)*map2.get(-t); return res; }