LeetCode Exercise 12:Minimum Size Subarray Sum
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题名:Minimum Size Subarray Sum
难度:Medium
题目描述:
代码如下:
class Solution {public: int minSubArrayLen(int s, vector<int>& nums) { int n = nums.size(), left = 0, right = 0; int sum = 0, minlen = INT_MAX; while (right < n) { do sum += nums[right++]; while (right < n && sum < s); while (left < right && sum - nums[left] >= s){ sum -= nums[left++]; } if (sum >= s){ minlen = min(minlen, right - left); } } return minlen == INT_MAX ? 0 : minlen; }};
个人体会:
由于题目要求的是最短的连续的子序列,所以在取舍时只要考虑左右两端即可。
hhh看了别人的解法,INT_MAX作为边界比较值挺好用的。
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