Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ≥ s. If there isn't one, return 0 instead.
Example:
Input:s = 7, nums = [2,3,1,2,4,3]
Output: 2 Explanation: the subarray[4,3]
has the minimal length under the problem constraint.
Follow up:
If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).
LINK:here
思路:
看到这种题首先想到的应该是滑动窗操作,我们使用两个指针记录窗口的左右边界,然后统计窗口中的值。使用滑动窗口最重要的一点是注意边界的更新条件。代码如下:
class Solution {
public:
int minSubArrayLen(int s, vector<int>& nums) {
int fast = 0, low = 0; //用于存放滑动窗口的左右边界
int minLen = INT_MAX; //最小长度
int sum = 0; //存放窗口中元素的和
while(fast < int(nums.size())){
sum += nums[fast++];
while(sum >= s){
minLen = min(minLen, fast-low);
if(minLen == 1) return 1;
sum -= nums[low++];
}
}
if(minLen == INT_MAX) return 0;
else return minLen;
}
};