Problem:
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.
Summary:
找到数组中和大于目标值s的最短子序列。
Solution:
用两个指针分别代表当前子序列的开始和结尾,若计算出的sum小于s,则end++,否则start--,每计算出一个更短的子序列,更新res值。
1 class Solution { 2 public: 3 int minSubArrayLen(int s, vector<int>& nums) { 4 int len = nums.size(); 5 int res = len + 1, start = 0, end = 0, sum = 0; 6 while (start < len && end < len) { 7 while (sum < s && end < len) { 8 sum += nums[end++]; 9 } 10 11 while (sum >= s && start <= end) { 12 res = min(res, end - start); 13 sum -= nums[start++]; 14 } 15 } 16 17 return res == len + 1 ? 0 : res; 18 } 19 };