LeetCode 0053. Maximum Subarray最大子序和【Easy】【Python】【动态规划】
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.
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。
进阶:
如果你已经实现复杂度为 O(n) 的解法,尝试使用更为精妙的分治法求解。
动态规划
找到 dp 递推公式。dp 等于每个位置的数字加上前面的 dp,当前面的 dp 是负数时就不要加了。
时间复杂度: O(len(nums))
空间复杂度: O(1)
class Solution(object):
def maxSubArray(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if not nums:
return 0
dp = 0
sum = -0xFFFFFFFF
for i in range(len(nums)):
dp = nums[i] + (dp if dp > 0 else 0) # if dp > 0: dp = nums[i] + dp, else: dp = nums[i]
sum = max(sum, dp)
return sum
LeetCode | 0053. Maximum Subarray最大子序和【Python】
原文:https://www.cnblogs.com/wonz/p/12341899.html