题目来源
https://leetcode.com/problems/maximum-subarray/description/
题目描述
Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.
Example:
Input: [-2,1,-3,4,-1,2,1,-5,4],
Output: 6
Explanation: [4,-1,2,1] has the largest sum = 6.
Follow up:
If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.
求出一个数组的最大连续子数组之和.
注意: 如果已经掌握了
的解法, 建议尝试用分治策略也做尝试.
解题思路
首先来看分治策略的解法. 这个问题似乎在算法导论一书的分治部分中是首先被讲到的经典问题. 大致思路可以这样表达:
依靠上面的分治式子, 或者说递归式子, 我们便可以很容易的写出解题代码了.
代码实现
class Solution:
def __init__(self):
nums = []
def maxSubArray(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
self.nums = nums
return self.findMax(0, len(nums))
def findMax(self, start, end):
if (start == end - 1 or start == end):
return self.nums[start]
mid = int((start + end) / 2)
leftMax = self.findMax(start, mid)
rightMax = self.findMax(mid, end)
midMax = self.findCross(start, end)
return max(leftMax, rightMax, midMax)
# Max subarray cross the middle element
def findCross(self, start, end):
if start == end - 1 or start == end:
return self.nums[start]
mid = int((start + end) / 2)
current = self.nums[mid]
ret = current
for i in range(mid + 1, end):
current += self.nums[i]
if current > ret:
ret = current
current = ret
for i in range(mid - 1, start - 1, -1):
current+=self.nums[i]
if current > ret:
ret = current
return ret
代码表现
解题思路
可以看到, 采用分治策略的解法表现可以说是相当差了. 其实这里最快的是采用动态规划策略的解法. 它的方程如下:
在实现的时候, 我们用一个lastMax来标记当前的子数组之和, 一旦它小于0就重新计数. 这是自然的, 这说明重新来过比加上前面的部分要好, 因为前面的部分已经小于0了. 一旦它大于我们维护的max(目标值, 即最大连续子数组之和), 我们便刷新max.
代码实现
import sys
class Solution:
def maxSubArray(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
size=len(nums)
if size<1:
return 0
ret,lastMax=nums[0],nums[0]
for i in range(1,size):
if lastMax<0:
lastMax=nums[i]
else:
lastMax+=nums[i]
if lastMax>ret:
ret=lastMax
return ret
代码表现
可以看到代码的速度有了极大的提高.