LeetCode 209. Minimum Size Subarray Sum

Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous 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.

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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.

https://leetcode.com/problems/minimum-size-subarray-sum/solution/
链接中给出的解答非常全面，暴力搜索会导致超时，用空间换时间会使得时间复杂度降低到n^2，通过使用二叉搜索可以使得复杂度进一步降低，这道题最好的方法是使用双指针，分别表示子序列的头和尾，遍历整个数组，sum表示之前所有元素之和，内部的while循环则每次去除下表最小的元素，逐步找到最短的符合要求子序列，最后结果返回ans

要点：贪心算法
伪代码如下：
1、初始化left = 0， sum = 0
2、for i = 0 to n 循环
sum累加
计算符合要求的子序列的长度，跟新ans

ps:最开始我也怀疑这么做能否找到最优解，后来仔细想了一下，这样做可以遍历所有的恰好满足要求的连续子数组，必定会找到最优解。

 1 int minSubArrayLen(int s, vector<int>& nums)
 2 {
 3     int n = nums.size();
 4     int ans = INT_MAX;
 5     int left = 0;
 6     int sum = 0;
 7     for (int i = 0; i < n; i++) {
 8         sum += nums[i];
 9         while (sum >= s) {
10             ans = min(ans, i + 1 - left);
11             sum -= nums[left++];
12         }
13     }
14     return (ans != INT_MAX) ? ans : 0;
15 }

时间复杂度：O(n)，每个元素最多被左右两个指针遍历两次。

空间复杂度：O(1)

二刷，笔写，2018-03-14，看着解释基本AC