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.

 

给出一个数组，找出数组中和大于等于s的最小长度。

 

1、计算sum，然后找出以每一位开头的和大于等于s的长度。

public class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        int len = nums.length;
        if (len == 0){
            return 0;
        }
        if (nums[0] >= s){
            return 1;
        }
        int i = 0, result = 0, sum = 0;
        for (; i < len; i++){
            sum += nums[i];
            if (sum >= s){
                break;
            }
        }
        if (i == len && sum < s){
            return 0;
        }
        i++;
        result = i;
        sum -= nums[0];
        for (int j = 1; j < len; j++){
            while (i < len && sum < s){
                sum += nums[i];
                i++;
            }
            if (sum < s){
                break;
            }
            result = Math.min(i - j, result);
            sum -= nums[j];
        }
        return result;
       
    }
}

 

2、还可以判断从0～len-1位，以每一位为结尾的最小的长度。

public class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        if (nums.length == 0){
            return 0;
        }
        int i = 0, j = 0, sum = 0, min = Integer.MAX_VALUE;
        
        while (j < nums.length){
            sum += nums[j++];
            
            while (sum >= s){
                min = Math.min(min, j - i);
                sum -= nums[i++];
            }
        }
        return min == Integer.MAX_VALUE ? 0 : min;

    }
}