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[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 1:
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
Tags: Hash Table
解读题意
给出ABCD4个数组,计算出有多少个(i, j, k, l)
能够使表达式 A[i] + B[j] + C[k] + D[l] =0
成立的组合
思路1
1. 定义两个map,分别用来存储key=A[i] + B[j],value=A[i] + B[j]出现次数
,key=C[k] + D[l],value=A[i] + B[j]出现次数
2. 遍历m1,因为x+(-x)=0
成立。x为map1的key,设-x即为其相反值,若m2.containsKey(-x)
成立,则符合表达式A[i] + B[j] + C[k] + D[l] =0
,结果集res += m1.get(x) * m2.get(-x)
;否则res += 0
如上图所示,ABCD数组分为A[-1,-1]、B[-1,1]、C[-1,1]、D[1,-1],则A[i]+B[j]
可能出现的值为0、-2,出现次数为2、2,为了方便,我们用(0,2),(-2,2)来表示。同理,C[k] + D[l]
可以表示为(0,2),(-2,1),(2,1)。当m1的entry为(0,2)时,m2的entry为(0,2)符合0+0=0
,所以此时res = 2*2=4,同理可以算出res=6
public class Solution {
public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
if (A.length != B.length || B.length != C.length || C.length != D.length
|| A.length == 0)
return 0;
Map<Integer, Integer> m1 = new HashMap<>();
Map<Integer, Integer> m2 = new HashMap<>();
// 默认数组ABCD有共同的length
int n = A.length;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
m1.put(A[i] + B[j], m1.getOrDefault(A[i] + B[j], 0) + 1);
m2.put(C[i] + D[j], m2.getOrDefault(C[i] + D[j], 0) + 1);
}
}
int res = 0;
for (int x : m1.keySet()) {
res += m2.containsKey(-x) ? m1.get(x) * m2.get(-x) : 0;
}
return res;
}
}
time complexity:O(n^2)。
leetCode汇总:https://blog.csdn.net/qingtian_1993/article/details/80588941
项目源码,欢迎star:https://github.com/mcrwayfun/java-leet-code