剑指offer--面试题31：连续子数组的最大和



题目描述

HZ偶尔会拿些专业问题来忽悠那些非计算机专业的同学。今天测试组开完会后,他又发话了:在古老的一维模式识别中,常常需要计算连续子向量的最大和,当向量全为正数的时候,问题很好解决。但是,如果向量中包含负数,是否应该包含某个负数,并期望旁边的正数会弥补它呢？例如:{6,-3,-2,7,-15,1,2,2},连续子向量的最大和为8(从第0个开始,到第3个为止)。你会不会被他忽悠住？

python实现：

# -*- coding:utf-8 -*-class Solution:         #法1：动态规划    def FindGreatestSumOfSubArray1(self, array):        # write code here        #dp[i]表示以array[i]为结尾的子数组(0, i)的最大和        #如果dp[i-1]>0, dp[i] = dp[i-1]+array[i]        #如果dp[i-1]<=0, dp[i] = array[i]        n = len(array)        if n==0:            return 0        dp = [0]*n        #初始化        dp[0] = array[0]        for i in range(1, n):            if dp[i-1]<=0:                dp[i] = array[i]            else:                dp[i] = dp[i-1] + array[i]        return max(dp)         #法2：分治    #参考：http://blog.csdn.net/a272846945/article/details/50822601    def FindGreatestSumOfSubArray(self, array):        n = len(array)        if n==0:            return 0        return self.devideAndConquer(array, 0, n-1)         def devideAndConquer(self, nums, low, high):        if low==high:            return nums[low]        mid = (low+high)/2        leftMaxSum = self.devideAndConquer(nums, low, mid)        rightMaxSum = self.devideAndConquer(nums, mid+1, high)        crossMaxSum = self.merge(nums, low, mid, high)        return max(leftMaxSum, rightMaxSum, crossMaxSum)         def merge(self, nums, low, mid, high):        continousLeftSum = nums[mid]        tmpLeftSum = nums[mid]        for i in range(mid-1, low-1, -1):            tmpLeftSum += nums[i]            continousLeftSum = max(continousLeftSum, tmpLeftSum)        continousRightSum = nums[mid+1]        tmpRightSum = nums[mid+1]        for i in range(mid+2, high+1):            tmpRightSum += nums[i]            continousRightSum = max(continousRightSum, tmpRightSum)        return continousLeftSum + continousRightSum