add 560. Subarray Sum Equals K 🍺

README.md

@@ -315,6 +315,7 @@ My LeetCode solutions with Chinese explanation. 我的LeetCode中文题解。
 | 509 |[Fibonacci Number](https://leetcode.com/problems/fibonacci-number/)|[C++](solutions/509.%20Fibonacci%20Number.md)|Easy|  |
 | 532 |[K-diff Pairs in an Array](https://leetcode.com/problems/k-diff-pairs-in-an-array)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/532.%20K-diff%20Pairs%20in%20an%20Array.md)|Easy|  |
 | 543 |[Diameter of Binary Tree](https://leetcode.com/problems/diameter-of-binary-tree/)|[C++](solutions/543.%20Diameter%20of%20Binary%20Tree.md)|Easy|  |
+| 560 |[Subarray Sum Equals K](https://leetcode.com/problems/subarray-sum-equals-k/)|[C++](solutions/560.%20Subarray%20Sum%20Equals%20K.md)|Medium|  |
 | 561 |[Array Partition I](https://leetcode.com/problems/array-partition-i)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/561.%20Array%20Partition%20I.md)|Easy|  |
 | 566 |[Reshape the Matrix](https://leetcode.com/problems/reshape-the-matrix)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/566.%20Reshape%20the%20Matrix.md)|Easy|  |
 | 572 |[Subtree of Another Tree](https://leetcode.com/problems/subtree-of-another-tree/)|[C++](solutions/572.%20Subtree%20of%20Another%20Tree.md)|Easy|  |

solutions/560. Subarray Sum Equals K.md

@@ -0,0 +1,30 @@
+# [560. Subarray Sum Equals K](https://leetcode.com/problems/subarray-sum-equals-k/)
+
+# 思路
+
+求和为k的连续子数组的个数。因为要求连续子数组，所以我们可以先计算出前缀和`preSum`，然后就可以很方便计算任意区间对应的连续子数组的和`sum[i~j] = preSum[j]-preSum[i]`。要和为k，所以我们可以用一个二重循环遍历前缀和数组`preSum`，但是复杂度为O(n^2)。
+
+更高效的做法是用一个hash表来记录所有前缀和与其出现次数的映射，这样假设当前位置的累加和为`cur_sum`，我们可以通过哈希表快速查找`cur_sum - k`出现的次数。注意hash表初始时要有一个0到1的映射。
+
+时间复杂度O(n)
+
+# C++
+``` C++
+class Solution {
+public:
+    int subarraySum(vector<int>& nums, int k) {
+        unordered_map<int, int>mp;
+        mp[0] = 1;
+        
+        int cur_sum = 0, res = 0;
+        for(int i = 0; i < nums.size(); i++){
+            cur_sum += nums[i];
+            auto it = mp.find(cur_sum - k);
+            if(it != mp.end()) res += it -> second;
+            mp[cur_sum]++;
+        }
+        return res;
+    }
+};
+```
+