## Description

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.


If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.

## Solutions

找到连续子序列最大加和的结果。

### 1. Array

# Time: O(n)
# Space: O(1)
class Solution:
def maxSubArray(self, nums: List[int]) -> int:
cur_sum = max_sum = nums[0]
for i in range(1, len(nums)):
if cur_sum <= 0:
cur_sum = nums[i]
else:
cur_sum += nums[i]
max_sum = max(max_sum, cur_sum)

return max_sum

# 202/202 cases passed (60 ms)
# Your runtime beats 94.35 % of python3 submissions
# Your memory usage beats 5.69 % of python3 submissions (14.6 MB)


### 2. DP

# Time: O(n)
# Space: O(n)
class Solution:
def maxSubArray(self, nums: List[int]) -> int:
n = len(nums)
dp = [0 for _ in range(n)]
dp[0] = nums[0]
max_sum = nums[0]
for i in range(1, n):
dp[i] = nums[i] + (dp[i-1] if dp[i-1] > 0 else 0)
max_sum = max(max_sum, dp[i])
return max_sum

# 202/202 cases passed (72 ms)
# Your runtime beats 28.46 % of python3 submissions
# Your memory usage beats 5.69 % of python3 submissions (14.7 MB)


滚动数组降低空间复杂度：

# Time: O(n)
# Space: O(1)
class Solution:
def maxSubArray(self, nums: List[int]) -> int:
n = len(nums)
dp_max_end_i = nums[0]
max_sum = nums[0]
for i in range(1, n):
dp_max_end_i = nums[i] + (dp_max_end_i if dp_max_end_i > 0 else 0)
max_sum = max(max_sum, dp_max_end_i)
return max_sum

# 202/202 cases passed (64 ms)
# Your runtime beats 82.31 % of python3 submissions
# Your memory usage beats 5.69 % of python3 submissions (14.7 MB)