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; } };