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
在给定的4个数组中各取1个数字，使4个数的和为0，返回找到的次数。
如果用暴力搜索Bruce Force，那么Time: O(n^4)，用时太长。
好的解法是用Hash map，建立一个map，将A，B中每两个数的和作为key，次数作为value写入map，然后查找C，D中每两个数和的相反数是否在map中，如果在，就把 map 中记录的次数累加到结果中。
Java：
public class Solution {
public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
int res = 0;
HashMap<Integer,Integer> ab = new HashMap<Integer,Integer>();
for(int a:A){
for(int b:B){
ab.put(a+b,ab.getOrDefault(a+b,0) + 1);
}
}
for(int c:C){
for(int d:D){
int part2 = c + d;
int part1 = - part2;
res += ab.getOrDefault(part1,0);
}
}
return res;
}
}
Python:
class Solution(object):
def fourSumCount(self, A, B, C, D):
ans = 0
cnt = collections.defaultdict(int)
for a in A:
for b in B:
cnt[a + b] += 1
for c in C:
for d in D:
ans += cnt[-(c + d)]
return ans
Python: 这个写法牛逼到了只用两行就搞定了
class Solution(object):
def fourSumCount(self, A, B, C, D):
A_B_sum = collections.Counter(a+b for a in A for b in B)
return sum(A_B_sum[-c-d] for c in C for d in D)
C++:
class Solution {
public:
int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
int res = 0;
unordered_map<int, int> m;
for (int i = 0; i < A.size(); ++i) {
for (int j = 0; j < B.size(); ++j) {
++m[A[i] + B[j]];
}
}
for (int i = 0; i < C.size(); ++i) {
for (int j = 0; j < D.size(); ++j) {
int target = -1 * (C[i] + D[j]);
res += m[target];
}
}
return res;
}
};
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