Minimum Size Subarray Sum
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2024-01-31 16:10:10
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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.
在一个数组中,找到一个长度最小的子数组,使它的和大于或等于给定的数字s。我们采用类似滑动窗口的方法来解决。从第一个元素开始,一直累加到和sum大于或等与s,然后我们记录这时的长度。接下来从开始的元素缩小这个窗口,缩小到sum小于s的时候,窗口向右移动,直到移动到数组末尾结束。代码如下:
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.
在一个数组中,找到一个长度最小的子数组,使它的和大于或等于给定的数字s。我们采用类似滑动窗口的方法来解决。从第一个元素开始,一直累加到和sum大于或等与s,然后我们记录这时的长度。接下来从开始的元素缩小这个窗口,缩小到sum小于s的时候,窗口向右移动,直到移动到数组末尾结束。代码如下:
public class Solution { public int minSubArrayLen(int s, int[] nums) { int start = 0; int sum = 0; int minLen = Integer.MAX_VALUE; for(int i = 0; i < nums.length; i++) { sum += nums[i]; if(sum >= s) { while(sum >= s) { minLen = Math.min(minLen, i - start + 1); sum -= nums[start]; start ++; } } } return minLen == Integer.MAX_VALUE ? 0 : minLen; } }
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