[LeetCode]Minimum Size Subarray Sum

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.

click to show more practice.

More practice:
If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).
思路：采用Window的思想，保证在窗内满足>=k的值，动态更新subarray的大小，找到所有subarray的最小值。扫描时动态更新边界。这样是O(N)的算法。

class Solution {
public:
int minSubArrayLen(int s, vector<int>& nums) {
int length = nums.size();
int ret = nums.size()+1;
int temp = 0;
int start = 0;
for(int i=0;i<length;++i){
if(temp<s){
temp += nums[i];
}
while(temp>=s){ //更新右边界
ret = (i-start+1)<ret?(i-start+1):ret;
temp = temp - nums[start];
start++; //更新左边界
}
}
if(ret == nums.size()+1) //判断是否存在结果
return 0;
return ret;
}
};