[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).

/*
* 双指针sta和end，大循环是每次end++，比较两个指针间所有数的和和target的大小，大于的话进小循环，不断sta++，知道小于，每次都更新res
* 两指针之间所有数的和的计算，一开始想到的是每次指针变动的时候都for循环进行相加，后来看了别人的方法，发现不用每次都重新加
* 只要当end++后cur加上当前值，sta++前cur减去当前值就行，有点像回溯
* 注意会有全部相加都小于的情况，最后要单独判断*/

public int minSubArrayLen(int s, int[] nums) {
        //双指针
        int sta = 0;
        int end = 0;
        //length
        int res = nums.length + 1;
        //当前的和
        int cur = 0;
        //大循环
        while (end < nums.length)
        {
            cur += nums[end];
            //小循环
            while (cur >= s)
            {
                res = Math.min(res,end - sta + 1);
                cur -= nums[sta];
                sta ++;
            }
                end++;
        }
        //特殊情况判断
        if (res == nums.length + 1)
            return 0;
        return res;
    }