if len(nums) == 1:
return nums[0]
else:
dp = [0] * len(nums)
dp[0] = nums[0]
for i in range(1,len(nums)):
if dp[i-1] > 0:
dp[i] = nums[i] + dp[i-1]
else:
dp[i] = nums[i]
return max(dp)
分治算法
n = len(nums)
#递归终止条件
if n == 1:
return nums[0]
else:
#递归计算左半边最大子序和
max_left = self.maxSubArray(nums[0:len(nums) // 2])
#递归计算右半边最大子序和
max_right = self.maxSubArray(nums[len(nums) // 2:len(nums)])
#计算中间的最大子序和,从右到左计算左边的最大子序和,从左到右计算右边的最大子序和,再相加
max_l = nums[len(nums) // 2 - 1]
tmp = 0
for i in range(len(nums) // 2 - 1, -1, -1):
tmp += nums[i]
max_l = max(tmp, max_l)
max_r = nums[len(nums) // 2]
tmp = 0
for i in range(len(nums) // 2, len(nums)):
tmp += nums[i]
max_r = max(tmp, max_r)
#返回三个中的最大值
return max(max_right,max_left,max_l+max_r)
# 假如这左、右两个能够拼接起来