[leetcode]Minimum Size Subarray Sum

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.

题意：给出包含n个正整数的数组，和一个定值s，在数组中找出最小的子数组串使其和大于或等于s。如果不存在这样的子数组则返回0

（子数组要连续）

解题思路：使用两个下标指针统计子数组的和， 当其和大于或等于s时， 第一个指针向前移动，找到其第一次小于s时的位置

如  1  2 3  4  5    i(初始化指向1)   j作为for循环中的下标向前移动  如s=11  当j指向5时， i向前移动直到子数组和小于s 计算长度

public class Solution {    public int minSubArrayLen(int s, int[] nums) {        if(nums == null || s < 0) return 0;        int i = 0;        int sum = 0;        int minLen = nums.length + 1;                for(int j = 0; j < nums.length; j++){            sum += nums[j];            while(sum >= s){                sum -= nums[i];                minLen = Math.min(minLen, j - i + 1);                i++;            }        }        if(minLen > nums.length) return 0;        return minLen;    }}