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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.
Following the thought of slide window, we apply two pointers here to solve the problem, one for left and one for right.
If current sum >= target, left++;
If current sum < target, right--;
Remember to check right index before getting nums[right], and sum value is nums[0] at beginning.
Time complexity O(n).
1 public class Solution { 2 public int minSubArrayLen(int s, int[] nums) { 3 if (nums == null || nums.length == 0) 4 return 0; 5 int length = nums.length, left = 0, right = 0, result = length, sum = nums[0]; 6 while (right < length) { 7 if (left == right) { 8 if (nums[left] >= s) { 9 return 1; 10 } else { 11 right++; 12 if (right < length) 13 sum += nums[right]; 14 else 15 break; 16 } 17 } else { 18 if (sum < s) { 19 right++; 20 if (right < length) 21 sum += nums[right]; 22 else 23 break; 24 } else { 25 result = Math.min(result, right - left + 1); 26 sum -= nums[left]; 27 left++; 28 } 29 } 30 } 31 if (left == 0 && sum < s) 32 result = 0; 33 return result; 34 } 35 }
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原文地址：http://www.cnblogs.com/ireneyanglan/p/4806040.html
