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.
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).
Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.
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解题思路：题目大概的意思是寻找最短的数组和大于等于sum, 用遍历的思想。第一个到滴n个， 答案（如4,3）一定是按顺序搜索到的。
两个指针sum和end标定。start从0開始
（1）我们从下标0開始搜索大于sum的，
（2）假设大于sum，则start往后一格。,end记住位置
（3）假设start-往后走。sum减小之前的start上的数，此时sum还大于s返回2，否则返回（1）
class Solution {
public:
int minSubArrayLen(int s, vector<int>& nums)
{
if(nums.size()==0)
return 0;
int start=0;
int end;
int numMin;
int AnswernumMin=INT_MAX;
int sum=0;
for (int i=0;i<nums.size();i++)
{
sum+=nums[i];
while(sum>=s)
{
end=i;
numMin=end-start+1;//length mean has 0 means 1,start=end;
if (numMin<AnswernumMin)
{
AnswernumMin=numMin;
}
sum=sum-nums[start];
start=start+1; //start forward one!
}
}
if (AnswernumMin==INT_MAX)
{
return 0;
}
return AnswernumMin;
}
};