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.题目里面没要求连续子数组...但是解题时得按照连续子数组来解,郁闷'''class Solution(object):    def minSubArrayLen(self, s, nums):        """        :type s: int        :type nums: List[int]        :rtype: int        非连续子数组,最少个数        """        length = len(nums)        if length == 0:            return 0        nums.sort()        index = length - 1        while index >= 0 and s > 0:            s = s - nums[index]            index -= 1        if s > 0:            return 0                return length - 1 -index    def minSubArrayLen(self, s, nums):        """        :type s: int        :type nums: List[int]        :rtype: int        leetcode答案,连续子数组最少个数        """        length = len(nums)        if length == 0:            return 0        sum = nums[0]        start,end = 0,1        min_len = length        tmp_len = 1        while end < length and start < end:            if sum < s:                sum += nums[end]                end += 1                tmp_len += 1             elif sum >= s:                if tmp_len < min_len:                    min_len = tmp_len                sum -= nums[start]                start += 1                tmp_len -= 1        if min_len == length and sum < s:#整个数组的和没有s大            return 0        while sum > s and (sum - nums[start]) >= s and start < length:            sum -= nums[start]            tmp_len -= 1            if tmp_len < min_len:                min_len = tmp_len            start += 1        return min_len