[leetcode] 209.Minimum Size Subarray Sum
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题目:
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.
题意:
给定一个数组包含n歌正数,给定一个整数,找出数组的子数组之和大于等于这个整数,找出所有的子数组的最短的那个。
思路:
这种题目还是使用两个指针进行操作,先让第二个指针从0的位置加到某个位置i,此时这些元素之和已经大于等于我们的给定值。然后此时尝试改变第一个指针,来缩小这个子数组,更新满足情况的子数组的长度,如果子数组和比给定值小了,我们就往后改变第二个指针,再次来扩大子数组。
总结起来就是:
- 如果和小于给定值,那么增大第二个指针,来扩大子数组之和。
- 如果子数组已经大于等于给定值了,那么增大第一个指针来缩小子数组的大小。
以上。
代码如下:
class Solution {public: int minSubArrayLen(int s, vector<int>& nums) { int size = nums.size(); if(size == 0)return 0; int first = -1, second = 0; int count = nums[0]; if(count == s)return 1; int min_length = INT_MAX; while(first < size && second < size) { if(count >= s) { min_length = min(min_length, second - first); first++; if(first == size)break; count -= nums[first]; } else { second++; if(second == size)break; count += nums[second]; } } return (min_length == INT_MAX)?0:min_length; }};
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