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