2019-08-28 长度最小的子数组

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