Minimum Size Subarray Sum

code

i前进，r指针不会退

要记录的值：

1） sum = 0;

2）指针；

3）需要ret的值（最长长度或者最短长度）

class Solution {
public:
    /**
     * @param nums: a vector of integers
     * @param s: an integer
     * @return: an integer representing the minimum size of subarray
     */
    int minimumSize(vector<int> &nums, int s) {
        // write your code here
        if(nums.size() == 0)
        {
            return -1;
        }

        int ret = INT_MAX;
        int r = 0;
        int sum = 0;

        for(int i = 0; i < nums.size(); i++)
        {
            while(r < nums.size())
            {
                if(sum + nums[r] < s)
                {
                    sum += nums[r];
                    r++;
                }
                else
                {
                    ret = min(ret, r-i+1);
                    break;
                }
            }
            sum -= nums[i];
        }

        return ret == INT_MAX?-1:ret;
    } 
};

class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        int n = nums.size();
        if(n == 0)
        {
            return 0;
        }

        int cur = 0;
        int cur_len = 0;
        int j = 0;
        int rst = INT_MAX;

        for(int i = 0; i<n; i++)
        {
            while(j < n)
            {
                if(cur >= s)
                {
                    break;
                }

                cur = cur + nums[j];
                j++;
                cur_len++;
            }

            if(cur >= s)
            {
                rst = min(rst, cur_len);
            }

            cur = cur - nums[i];
            cur_len--;
        }

        return rst==INT_MAX?0:rst;
    }
};