51 leetcode - Minimum Size Subarray Sum

#!/usr/bin/python
# -*- coding: utf-8 -*-
'''
Minimum Size Subarray Sum 
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.
題目裡面沒要求連續子數組...但是解題時得按照連續子數組來解,郁悶
'''
class Solution(object):
    def minSubArrayLen(self, s, nums):
        """
        :type s: int
        :type nums: List[int]
        :rtype: int
        非連續子數組,最少個數
        """
        length = len(nums)
        if length == :
            return 

        nums.sort()
        index = length - 
        while index >=  and s > :
            s = s - nums[index]
            index -= 

        if s > :
            return         
        return length -  -index

    def minSubArrayLen(self, s, nums):
        """
        :type s: int
        :type nums: List[int]
        :rtype: int
        leetcode答案,連續子數組最少個數
        """
        length = len(nums)
        if length == :
            return 

        sum = nums[]
        start,end = ,
        min_len = length
        tmp_len = 

        while end < length and start < end:
            if sum < s:
                sum += nums[end]
                end += 
                tmp_len +=  
            elif sum >= s:
                if tmp_len < min_len:
                    min_len = tmp_len
                sum -= nums[start]
                start += 
                tmp_len -= 

        if min_len == length and sum < s:#整個數組的和沒有s大
            return 

        while sum > s and (sum - nums[start]) >= s and start < length:
            sum -= nums[start]
            tmp_len -= 
            if tmp_len < min_len:
                min_len = tmp_len
            start += 

        return min_len