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).

题意：获取最短子串的长度，使其总和大于等于给定的值s

 

    public int minSubArrayLen(int s, int[] nums) {
//        定义两个指针left和i，分别记录子数组的左右的边界位置，每次i右移1位，就将结果累加到sum中。
//        找出left到i范围内sum大于s时最小的长度（i-left+1），然后将left右移，同时在sum中去除nums[left]的值
        int res = Integer.MAX_VALUE, left = 0, sum = 0;
        for (int i = 0; i < nums.length; ++i) {
            sum += nums[i];
            while (left <= i && sum >= s) {
                res = Math.min(res, i - left + 1);
                sum -= nums[left++];
            }
        }
        return res == Integer.MAX_VALUE ? 0 : res;        
    }