给定一个含有 n 个正整数的数组和一个正整数 s ，找出该数组中满足其和 ≥ s 的长度最小的连续子数组。如果不存在符合条件的连续子数组，返回 0。
示例:
输入: s = 7, nums = [2,3,1,2,4,3]
输出: 2
解释: 子数组 [4,3] 是该条件下的长度最小的连续子数组。
进阶:
如果你已经完成了O(n) 时间复杂度的解法, 请尝试 O(n log n) 时间复杂度的解法。
Python
class Solution:
# def minSubArrayLen(self, s: int, nums: List[int]) -> int:
# class Solution:
def minSubArrayLen(self, s, nums):
"""
:type s: int
:type nums: List[int]
:rtype: int
"""
l = 0
r = 0
sum_all = 0
nums_len = len(nums)
minLength = nums_len + 1
while l < nums_len:
if r < nums_len and sum_all < s:
sum_all += nums[r]
r += 1
else:
sum_all -= nums[l]
l += 1
if sum_all >= s:
minLength = min(minLength, r - l)
if minLength == nums_len + 1:
return 0
return minLength
C++ O(nlgn)
class Solution {
public:
int minSubArrayLen(int s, vector<int>& nums) {
int minLen = nums.size()+1;
int len = nums.size();
int sum = 0;
int i=0,j=0;
while(i < len){
if(j <len && sum < s){
sum += nums[j];
++j;
}else{
sum -= nums[i];
++i;
}
if(sum >= s){
minLen = minLen>j-i?j-i:minLen;
}
}
if(minLen == len + 1){
return 0;
}
return minLen;
}
};
Python
class Solution:
def _windowEx(self, nums, size, s):
sum = 0
for i, _ in enumerate(nums):
if i >= size:
sum -= nums[i - size]
sum += nums[i]
if sum >= s:
return True
return False
def minSubArrayLen(self, s, nums):
"""
:type s: int
:type nums: List[int]
:rtype: int
"""
l = 1
r = len(nums)
result = 0
while l <= r:
mid = l + (r - l)//2
if self._windowEx(nums, mid, s):
r = mid - 1
result = mid
else:
l = mid + 1
return result
