LeetCode454 4Sum II

Posted 2020-09-11 Vincent丶

题目：

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

　　

题解：

　　嗯，老规矩，直接暴力解，一共四个循环。

Solution 1 (TLE)

 1 class Solution {
 2 public:
 3     int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
 4         sort(A.begin(), A.end()); 
 5         sort(B.begin(), B.end());
 6         sort(C.begin(), C.end());
 7         sort(D.begin(), D.end());  
 8         int n = A.size();
 9         int cnt = 0;
10         for(int i=0; i<n; i++) {
11             for(int j=0; j<n; j++) {
12                 for(int k=0; k<n; k++) {
13                     for(int l=0; l<n; l++) {
14                         int sum = A[i] + B[j] + C[k] + D[l];
15                         if(sum==0) {
16                             cnt++;
17                         }
18                         else if(sum<0) continue;
19                         else break;
20                     }
21                 }
22             }
23         }
24         return cnt;
25     }
26 };

　　不出意外，TLE了，哈哈。那怎么降低时间复杂度呢？我基本上第一时间想到的就是双指针和map(set)，这个明显双指针一般用于单个数组的操作，因此可以考虑使用map。思路：使用两个循环，循环一：建立两个map，一个是AB两数组元素和与出现次数的map，另一个为CD两数组元素和与出现次数的map，循环二：遍历map1，map2使用find函数来查找是否存在遍历元素的相反数。

Solution 2 (706ms)

 1 class Solution {
 2 public:
 3     int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
 4         unordered_map<int,int> m1, m2;
 5         int result = 0;
 6         int n = A.size();
 7         
 8         for(int i = 0; i < n; ++i) {
 9             for(int j = 0; j < n; ++j) {
10                   ++m1[A[i] + B[j]];
11                   ++m2[C[i] + D[j]];
12               }
13         }
14     /*    for(auto it = m1.begin(); it != m1.end(); ++it) {
15             if(m2.find(-it->first) != m2.end())
16                 result += it->second * m2[-it->first];
17         }  */
18         for (auto p : m1) result += p.second * m2[-p.first];
19         return result;
20     }
21 };

　　另一个解法也是利用两个循环，不过只使用一个map，建立AB两数组元素和与出现次数的map后，对CD遍历元素求和得opp，result += m[-opp]，若-opp不存在，则m[-opp]为0，若存在，则result加上其出现的次数。

Solution 3 (449ms)

 1 class Solution {
 2 public:
 3     int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
 4         int result = 0, n = A.size();
 5         unordered_map<int, int> m;
 6         for (int i=0; i<n; ++i) {
 7             for (int j=0; j<n; ++j) {
 8                 ++m[A[i] + B[j]];
 9             }
10         }
11         for (int i=0; i<n; ++i) {
12             for (int j=0; j<n; ++j) {
13                 int opp = -1 * (C[i] + D[j]);
14                 result += m[opp];
15             }
16         }
17         return result;
18     }
19 };