LeetCode--053--最大子序和

问题描述:

给定一个整数数组 nums ，找到一个具有最大和的连续子数组（子数组最少包含一个元素），返回其最大和。

示例:

输入: [-2,1,-3,4,-1,2,1,-5,4],
输出: 6
解释: 连续子数组 [4,-1,2,1] 的和最大，为 6。

时间超限：  暴力穷举

 class Solution(object):
     def maxSubArray(self, nums):
         """
         :type nums: List[int]
         :rtype: int
         """

         max = nums[0]
         if len(nums) == 1:
             return nums[0]
         res = 0
         for step in range(len(nums)):#step 控制连续加的个数
             for i in range(len(nums)-step):#i控制从第几个开始加
                 for j in range(step+1):
                     res += nums[i]
                     i += 1
                 if res > max:
                     max = res
                 res = 0
         return max

方法1：当前值的大小与前面的值之和比较，若当前值更大，则取当前值，舍弃前面的值之和

 class Solution(object):
     def maxSubArray(self, nums):
         """
         :type nums: List[int]
         :rtype: int
         """
         if len(nums) == 0:
             return 0
         preSum = maxSum = nums[0]
         for i in xrange(1, len(nums)):
             preSum = max(preSum + nums[i], nums[i])
             maxSum = max(maxSum, preSum)
         return maxSum

方法2：（分治法）对半分，求左边最大，右边最大，以及边界最大 ，返回最大值

 class Solution(object):
     def maxSubArray(self, nums):
         """
         :type nums: List[int]
         :rtype: int
         """
         def maxSum(alist, left, right):
             #递归返回条件
             if left >= right:
                 return alist[left]  #return原值，不是return 0

             middle = (left + right) // 2 #记得打括号，去（TiMe）调了老半天
             leftMax = maxSum(alist, left, middle)
             rightMax = maxSum(alist, middle+1, right)

             #求左边界最大值
             leftBoardSum, leftBoardMax = 0, alist[middle]
             for i in range(middle, left-1,-1): #左段最右端没有取到middle
                 leftBoardSum += alist[i]
                 if leftBoardSum > leftBoardMax:
                     leftBoardMax = leftBoardSum

             #求右边界最大值
             rightBoardSum, rightBoardMax = 0, alist[middle+1]
             for j in range(middle+1, right+1): #右段最右端取到了right
                 rightBoardSum += alist[j]
                 if rightBoardSum > rightBoardMax:
                     rightBoardMax = rightBoardSum

             #边界最大值
             boardMax = leftBoardMax + rightBoardMax

             return max(leftMax, rightMax, boardMax)

         if nums == []:
             return 0

         left = 0
         right = len(nums)-1
         res = maxSum(nums, left, right)   #left,right为左右下标
         return res

2018-07-24 11:11:59