[leetcode]Minimum Size Subarray Sum

题目描述如下：

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.

题目大意就是寻找到最短的连续子串和大于等于给定的数。

此类题目之前有接触过，一般用的是滑动窗口的方法，即两个指针相互行进，扫描一遍数组后得出答案。

第一版代码如下:

public class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        int res = Integer.MAX_VALUE;
        int arrLen = nums.length;
        int first = 0, second = 0;
        int currSum = 0;
        boolean flag = true;
        while(first < arrLen){
            if(flag){
                currSum += nums[first];
                flag = false;
            }
            if(currSum >= s){
                if(res > first - second + 1){
                    res = first - second + 1;
                }
                currSum -= nums[second];
                second ++;
            }else{
                first ++;
                flag = true;
            }
        }
            return res;
    }
}

这样的做法超时，也就是说每次循环若只移动两个指针中的其一一格是不够的。于是进行优化，即在一次循环中，让前后指针在给定的范围内不断移动，这样就可以减少判断的次数。修改代码如下：

public class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        if(nums.length == 0) return 0;
        int res = Integer.MAX_VALUE;
        int arrLen = nums.length;
        int first = 0, second = 0;
        int currSum = 0;
        while(first < arrLen && second < arrLen){
                while(currSum < s && first < arrLen){
                    currSum += nums[first];
                    first++;
                }
                while(currSum >= s && second < first){
                    if(res > first - second){
                        res = first - second;
                    }
                    currSum -= nums[second];
                    second ++;
                }
            }
 return res;
    }
}

但是光做到这样还是不行，还需要对特殊条件进行判断：

1、输入的数组为空时，返回的值是0；

2、当数组无法满足给定数值时，返回的也是0；

加上对特殊条件的鉴别，最终代码如下：

public class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        if(nums.length == 0) return 0;
        int res = Integer.MAX_VALUE;
        int arrLen = nums.length;
        int first = 0, second = 0;
        int currSum = 0;
        while(first < arrLen && second < arrLen){
                while(currSum < s && first < arrLen){
                    currSum += nums[first];
                    first++;
                }
                while(currSum >= s && second < first){
                    if(res > first - second){
                        res = first - second;
                    }
                    currSum -= nums[second];
                    second ++;
                }
            }
 return res == Integer.MAX_VALUE ? 0 : res;
    }
}

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