lintcode 中等题：Minimum Size Subarray 和大于S的最小子数组

public class Solution {
    /**
     * @param nums: an array of integers
     * @param s: an integer
     * @return: an integer representing the minimum size of subarray
     */
    public int minimumSize(int[] nums, int s) {
        // write your code here
        if(nums.length == 0 || nums == null)
            return -1;
        int slow = 0;
        int fast = 0;
        int res = Integer.MAX_VALUE;
        int sum = 0;
        while(fast < nums.length){
            while( fast < nums.length && sum < s){
                sum += nums[fast];
                fast += 1;
                // System.out.println(sum+" "+fast);
            }
            // System.out.println();
            while(sum>=s){
                res = Math.min(res,fast - slow);
                sum -=nums[slow];
                slow +=1;
            }
        }
        return res==Integer.MAX_VALUE?-1:res;
    }
}

class Solution:
     # @param nums: a list of integers
     # @param s: an integer
     # @return: an integer representing the minimum size of subarray
    def minimumSize(self, nums, s):
        # write your code here
        if s == None or len(nums) == 0:
            return -1;
        lens = len(nums)
        slow = 0
        fast = 0
        sum = 0
        res = lens+1
        while fast < lens:
            while fast < lens and sum < s:
                sum += nums[fast]
                fast +=1
            while slow < fast and sum>= s:
                res = min(res,fast - slow)
                sum -= nums[slow]
                slow +=1
        if res ==lens+1:
            return -1
        else:
            return res