Minimum Size Subarray Sum
2016-05-29 21:30
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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.
这道题给定了我们一个数字,让我们求子数组之和大于等于给定值的最小长度,跟之前那道 Maximum Subarray 最大子数组有些类似,并且题目中要求我们实现O(n)和O(nlgn)两种解法,那么我们先来看O(n)的解法,我们需要定义两个指针left和right,分别记录子数组的左右的边界位置,然后我们让right向右移,直到子数组和大于等于给定值或者right达到数组末尾,此时我们更新最短距离,并且将left像右移一位,然后再sum中减去移去的值,然后重复上面的步骤,直到right到达末尾,且left到达临界位置,即要么到达边界,要么再往右移动,和就会小于给定值。代码如下:
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.
这道题给定了我们一个数字,让我们求子数组之和大于等于给定值的最小长度,跟之前那道 Maximum Subarray 最大子数组有些类似,并且题目中要求我们实现O(n)和O(nlgn)两种解法,那么我们先来看O(n)的解法,我们需要定义两个指针left和right,分别记录子数组的左右的边界位置,然后我们让right向右移,直到子数组和大于等于给定值或者right达到数组末尾,此时我们更新最短距离,并且将left像右移一位,然后再sum中减去移去的值,然后重复上面的步骤,直到right到达末尾,且left到达临界位置,即要么到达边界,要么再往右移动,和就会小于给定值。代码如下:
class Solution { public: int minSubArrayLen(int s, vector<int>& nums) { if(nums.empty()) return 0; int left = 0; int right = 0; int sum = 0; int len = nums.size(); int res = INT_MAX; while(right < len){ while(sum <s && right<len){ sum += nums[right++]; } if(sum>=s){ res = min (res,right-left); sum -= nums[left++]; } } if(res == INT_MAX) return 0; else return res; } };
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