209. Minimum Size Subarray Sum

Given an array of positive integers nums and a positive integer target, return the minimal length of a

subarray

whose sum is greater than or equal to target. If there is no such subarray, return 0 instead.

Example 1:

Input: target = 7, nums = [2,3,1,2,4,3]
Output: 2
Explanation: The subarray [4,3] has the minimal length under the problem constraint.

Example 2:

Input: target = 4, nums = [1,4,4]
Output: 1

Example 3:

Input: target = 11, nums = [1,1,1,1,1,1,1,1]
Output: 0

Constraints:

• 1 <= target <= 109
• 1 <= nums.length <= 105
• 1 <= nums[i] <= 104

Follow up: If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log(n)).

class Solution {
public:
    int minSubArrayLen(int target, vector<int>& nums) {
        int i=0;
        int result=INT32_MAX;
        int sum=0;
        int subLength=0;
        for(int j=0;j<nums.size();j++){
            sum+=nums[j];
            while(sum>=target){
                subLength=j-i+1;
                result=min(result,subLength);
                sum=sum-nums[i++];
            }
        }
        return result==INT32_MAX?0:result;
    }
};

注意：滑动窗口

        1.i表示起始位置，j表示终止位置

        2.sum>=target要用while，而不是if

        3.return的时候要多加上一个二元判断