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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.
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.
sliding window，两个pointers，start和end，当当前window值大于等于s是，计算长度若更短则记下来，接着收缩window直到值小于s。
时间复杂度O(n)
public class Solution {
public int minSubArrayLen(int s, int[] nums) {
int start = 0;
int end = 0;
int minLength = Integer.MAX_VALUE;
int sum = 0;
int length = nums.length;
while (end < length) {
while (sum < s && end < length) {
sum += nums[end++];
}
while (sum >= s && start < length) {
if (end - start < minLength) {
minLength = end - start;
}
sum -= nums[start++];
}
}
return minLength == Integer.MAX_VALUE ? 0 : minLength;
}
}
209. Minimum Size Subarray Sum
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原文地址：http://www.cnblogs.com/shini/p/4496290.html
