LeetCode:Minimum Size Subarray Sum

Minimum Size Subarray Sum

Total Accepted: 41802 Total Submissions: 153088 Difficulty: Medium

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.

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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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思路：

使用两个指针维持“最小”范围。时间复杂度是O(n)。

java code:

public class Solution {    public int minSubArrayLen(int s, int[] nums) {                int sum = 0;        int minLen = Integer.MAX_VALUE;        int end, start = 0; // 两个指针，维持最小范围        for(end = 0; end < nums.length; end++) {            sum += nums[end];            while(sum >= s) {                minLen = Math.min(minLen, end - start + 1);                sum -= nums[start++];            }        }        return minLen == Integer.MAX_VALUE ? 0 : minLen;    }}

上面说了时间复杂度是O(n)。虽然for中又套了个while，假设最坏的情况上sum每执行一次“+”，都要进行while循环，而且每次都要做“-”操作，这样也是执行1次“-”，因此总的时间是O(2n)。