add 454. 4Sum II 🍺

README.md

@@ -291,6 +291,7 @@ My LeetCode solutions with Chinese explanation. 我的LeetCode中文题解。
 | 447 |[Number of Boomerangs](https://leetcode.com/problems/number-of-boomerangs)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/447.%20Number%20of%20Boomerangs.md)|Easy|  |
 | 448 |[Find All Numbers Disappeared in an Array](https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/448.%20Find%20All%20Numbers%20Disappeared%20in%20an%20Array.md)|Easy|  |
 | 453 |[Minimum Moves to Equal Array Elements](https://leetcode.com/problems/minimum-moves-to-equal-array-elements)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/453.%20Minimum%20Moves%20to%20Equal%20Array%20Elements.md)|Easy|  |
+| 454 |[4Sum II](https://leetcode.com/problems/4sum-ii/)|[C++](solutions/454.%204Sum%20II.md)|Medium|  |
 | 455 |[Assign Cookies](https://leetcode.com/problems/assign-cookies)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/455.%20Assign%20Cookies.md)|Easy|  |
 | 459 |[Repeated Substring Pattern](https://leetcode.com/problems/repeated-substring-pattern)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/459.%20Repeated%20Substring%20Pattern.md)|Easy|  |
 | 461 |[Hamming Distance](https://leetcode.com/problems/hamming-distance)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/461.%20Hamming%20Distance.md)|Easy|  |

solutions/454. 4Sum II.md

@@ -0,0 +1,37 @@
+# [454. 4Sum II](https://leetcode.com/problems/4sum-ii/)
+
+# 思路
+
+从四个数组中各取一个数字，使其和为0，问有多少种取法。如果暴力四重循环的话复杂度为O(n^4)，是不可接受的。 我们可以采用hashmap将前两个数组的元素的和存放起来，这样就可以将复杂度降至O(n^2)。
+
+注意如果某个元素不在hashmap中，如果我们访问了它，那它会被插入到hashmap中，例
+``` C++
+unordered_map<int, int>mp;
+int a = mp[2];
+if(mp.count(2)) cout << "2 is in mp!" ; // 会输出
+```
+
+
+# C++
+``` 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>abSum;
+        for(int a: A)
+            for(int b: B)
+                abSum[a+b]++;
+        
+        for(int c: C)
+            for(int d: D){
+                // 以下两行比直接 res += abSum[-c-d] 要快,
+                // 因为如果 -c-d 不在abSum中, 访问abSum[-c-d]会将-c-d插入abSum中
+                auto it = abSum.find(-c-d);
+                if(it != abSum.end()) res += it -> second;
+            }
+
+        return res;
+    }
+};
+```