New Problem "Minimum Size Subarray Sum"

README.md

@@ -7,6 +7,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|193|[Minimum Size Subarray Sum](https://leetcode.com/problems/minimum-size-subarray-sum/)| [C++](./algorithms/minimumSizeSubarraySum/MinimumSizeSubarraySum.cpp)|Medium|
 |192|[Implement Trie (Prefix Tree)](https://leetcode.com/problems/implement-trie-prefix-tree/)| [C++](./algorithms/implementTriePrefixTree/ImplementTriePrefixTree.cpp)|Medium|
 |191|[Course Schedule](https://leetcode.com/problems/course-schedule/)| [C++](./algorithms/courseSchedule/CourseSchedule.cpp)|Medium|
 |190|[Reverse Linked List](https://leetcode.com/problems/reverse-linked-list/)| [C++](./algorithms/reverseLinkedList/reverseLinkedList.cpp)|Easy|

algorithms/minimumSizeSubarraySum/MinimumSizeSubarraySum.cpp

@@ -0,0 +1,51 @@
+// Source : https://leetcode.com/problems/minimum-size-subarray-sum/
+// Author : Hao Chen
+// Date   : 2015-06-09
+
+/********************************************************************************** 
+ * 
+ * Given an array of n positive integers and a positive integer s, find the minimal length of a subarray of which the sum ≥ s. If there isn't one, return 0 instead.
+ * 
+ * For example, given the array [2,3,1,2,4,3] and s = 7,
+ * the subarray [4,3] has the minimal length under the problem constraint.
+ * 
+ * click to show more practice.
+ * 
+ * More practice:
+ * 
+ * If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).
+ * 
+ * Credits:Special thanks to @Freezen for adding this problem and creating all test cases.
+ *               
+ **********************************************************************************/
+
+
+class Solution {
+public:
+    int minSubArrayLen(int s, vector<int>& nums) {
+        int min = nums.size();
+        int begin=0, end=0;
+        int sum = 0;
+        bool has = false;
+        
+        for (int i=0; i<nums.size(); i++, end++){
+            
+            sum += nums[i];
+            
+            while (sum >= s && begin <= end) {
+                //the 1 is minial length, just return
+                if (begin == end) return 1;
+                
+                if (end-begin+1 < min) {
+                    min = end - begin + 1;
+                    has = true;
+                }
+                //move the begin
+                sum -= nums[begin++];
+            }
+ 
+        }
+        
+        return has ? min : 0; 
+    }
+};