209 Minimum Size Subarray Sum

题目链接：https://leetcode.com/problems/minimum-size-subarray-sum/

题目：

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.

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解题思路：
这题的考点是由两个指针组成的滑动窗口。
We can use 2 points to mark the left and right boundaries of the sliding window. When the sum is greater than the target, shift the left pointer; when the sum is less than the target, shift the right pointer.
使用两个指针来标记滑动窗口的左右界限。
当总和大于等于目标值时，移动左指针；当总和小于目标值时，滑动右指针。

参考链接：http://www.programcreek.com/2014/05/leetcode-minimum-size-subarray-sum-java/

代码实现：

public class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        if(nums == null || nums.length == 0)
            return 0;
        int i = 0;
        int j = 0;
        int sum = nums[0];
        int res = nums.length;
        while(j < nums.length) {
            if(i == j) { // 单个元素足够大
                if(sum >= s)
                    return 1;
                else {
                    j ++;
                    if(j < nums.length) {
                        sum += nums[j];   
                    } else {
                        return res;
                    }
                }
            } else { 
                if(sum >= s) { // 和足够大，移动左指针
                res = Math.min(j - i + 1, res);
                    sum -= nums[i];
                    i ++;
                } else { // 和不够大，移动右指针
                    j ++;
                    if(j < nums.length) {
                        sum += nums[j];   
                    } else {
                        if(i == 0) // 所有元素的和都没有达到 s
                            return 0;
                        else
                            return res;
                    }
                }
            }
        }
        return res;
    }
}

14 / 14 test cases passed.
Status: Accepted
Runtime: 1 ms