Solution4.java

// 209. Minimum Size Subarray Sum
// https://leetcode.com/problems/minimum-size-subarray-sum/description/
//
// 另外一个滑动窗口的实现, 仅供参考
// 时间复杂度: O(n)
// 空间复杂度: O(1)
public class Solution4 {

 public int minSubArrayLen(int s, int[] nums) {

 if(s <= 0 || nums == null)
 throw new IllegalArgumentException("Illigal Arguments");

 int l = 0 , r = -1; // [l...r]为我们的窗口
 int sum = 0;
 int res = nums.length + 1;

 while(r + 1 < nums.length){ // 窗口的右边界无法继续扩展了, 则循环继续

 while(r + 1 < nums.length && sum < s)
 sum += nums[++r];

 if(sum >= s)
 res = Math.min(res, r - l + 1);

 while(l < nums.length && sum >= s){
 sum -= nums[l++];
 if(sum >= s)
 res = Math.min(res, r - l + 1);
 }
 }

 if(res == nums.length + 1)
 return 0;
 return res;
 }

 public static void main(String[] args) {

 int[] nums = {2, 3, 1, 2, 4, 3};
 int s = 7;
 System.out.println((new Solution4()).minSubArrayLen(s, nums));
 }
}