#406 Minimum Size Subarray Sum

题目描述：

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 -1 instead.

Have you met this question in a real interview? 

Yes

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.

Challenge 

If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).

题目思路：

这题还是用two pointers的方法，既然要算subarray的和，那么用一个presum去记录每个index之前的和，可以省去重复计算subarray和的麻烦。用2个pointers，都从头开始，如果subsum >= s，那么l向右移看看有没有更短的答案；反之，r向右移拉长subarray，看看有没有符合条件的答案。

Mycode（AC = 298ms）：

class Solution {public:    /**     * @param nums: a vector of integers     * @param s: an integer     * @return: an integer representing the minimum size of subarray     */    int minimumSize(vector<int> &nums, int s) {        // write your code here        if (nums.size() == 0) return -1;                // 0,2,5,6,8,12,15        // get presum        vector<int> presum(nums.size() + 1, 0);        for (int i = 1; i <= nums.size(); i++) {            presum[i] = presum[i - 1] + nums[i - 1];        }                // two pointers to find the min length        int l = 0, r = 1, minlen = INT_MAX;        while (r <= nums.size()) {            // if sum >= s, then increase l to find            // shorter answer            if (presum[r] - presum[l] >= s) {                minlen = min(minlen, r - l);                l++;            }            // if sum < s, then increase r to find answer            else {                r++;            }        }                return minlen == INT_MAX? -1 : minlen;    }};