解法一
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。
def maxSubArray(nums):
length = len(nums)
for i in range(1, length):
# 当前值的大小与前面的值之和比较,若当前值更大,则取当前值,舍弃前面的值之和
subMaxSum = max(nums[i] + nums[i - 1], nums[i])
nums[i] = subMaxSum # 将当前和最大的赋给nums[i],新的nums存储的为和值
return max(nums)
alist=[-2,1,-3,4,-1,2,1,-5,4]
print(maxSubArray(alist))
解法二
def maxSubArray( nums):
"""
:type nums: List[int]
:rtype: int
"""
# onesum维护当前的和
onesum = 0
maxsum = nums[0]
for i in range(len(nums)):
onesum += nums[i]
maxsum = max(maxsum, onesum)
# 出现onesum<0的情况,就设为0,重新累积和
if onesum < 0:
onesum = 0
return maxsum
alist=[-2,1,-3,4,-1,2,1,-5,4]
print(alist)
print(maxSubArray(alist))