Description
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).
分析
题目的意思是:求连续子数组只和等于s的最小长度。
- 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
class Solution {
public:
int minSubArrayLen(int s, vector<int>& nums) {
if(nums.empty()){
return 0;
}
int i=0;
int j=i+1;
int sum=nums[i];
int m=nums.size();
int len=m+1;
while(j<m){
while(sum<s&&j<m){
sum+=nums[j];
j++;
}
while(sum>=s){
len=min(len,j-i);
sum-=nums[i];
i++;
}
}
return len==m+1 ? 0:len;
}
};
代码二 (python)
- 用了二分法来缩小空间,(腾讯面试被问到,当时觉得自己没写好,所以结束后重新写一下,发现效率没有双指针高)
class Solution:
def minSubArrayLen(self, s: int, nums: List[int]) -> int:
n=len(nums)
sums=[0]*(n+1)
for i in range(1,n+1):
sums[i]=sums[i-1]+nums[i-1]
min_len=float('inf')
for i in range(0,n):
l=i
r=n-1
while(l<=r):
mid=l+(r-l)//2
t=sums[mid+1]-sums[i]
if(t<s):
l=mid+1
elif(t==s):
min_len=min(min_len,mid+1-i)
break
else:
r=mid-1
min_len=min(min_len,mid+1-i)
if(min_len==float('inf')):
return 0
return min_len
参考文献
[LeetCode] Minimum Size Subarray Sum 最短子数组之和