Minimum Size Subarray Sum

最新推荐文章于 2020-07-31 16:43:55 发布

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最新推荐文章于 2020-07-31 16:43:55 发布

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本文介绍了一种寻找数组中最短连续子数组的方法，该子数组的和至少为给定值s。通过双指针技术实现O(n)的时间复杂度，并提供了一个C++示例解决方案。

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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).

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;
		}
};