動機
解一解
Problem
Given an integer array nums
, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.
A subarray is a contiguous part of an array.
Example 1:
Input: nums = [-2,1,-3,4,-1,2,1,-5,4]Output: 6Explanation: [4,-1,2,1] has the largest sum = 6.
Example 2:
Input: nums = [1]Output: 1
Example 3:
Input: nums = [5,4,-1,7,8]Output: 23
Constraints:
1 <= nums.length <= 3 * 104
-105 <= nums[i] <= 105
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.
Sol
def helper(l):
if len(l) == 0:
return 0
ret = tmp = l[0]
for n in l[1:]:
tmp += n
ret = max(ret, tmp)
return ret
class Solution:
def maxSubArray(self, l: List[int]) -> int:
if not l:
return 0
if len(l) == 1:
return l[0]
else:
if len(l) % 2 == 0:
mid = len(l) // 2
onlyLeft = self.maxSubArray(l[:mid])
onlyRight = self.maxSubArray(l[mid:])
center = 0
left = helper(l[:mid][::-1])
right = helper(l[mid:])
else:
mid = len(l) // 2
onlyLeft = self.maxSubArray(l[:mid+1])
onlyRight = self.maxSubArray(l[mid:])
center = l[mid]
left = helper(l[:mid][::-1])
right = helper(l[mid+1:])
return max([center + left + right, onlyLeft, onlyRight])
Sol2
class Solution:
def maxSubArray(self, nums: List[int]) -> int:
ret = tmp = nums[0]
for n in nums[1:]:
tmp = max(n, n+tmp)
ret = max(ret, tmp)
return ret