解法一
給定一個整數數組 nums ,找到一個具有最大和的連續子數組(子數組最少包含一個元素),傳回其最大和。
def maxSubArray(nums):
length = len(nums)
for i in range(1, length):
# 目前值的大小與前面的值之和比較,若目前值更大,則取目前值,舍棄前面的值之和
subMaxSum = max(nums[i] + nums[i - 1], nums[i])
nums[i] = subMaxSum # 将目前和最大的賦給nums[i],新的nums存儲的為和值
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維護目前的和
onesum = 0
maxsum = nums[0]
for i in range(len(nums)):
onesum += nums[i]
maxsum = max(maxsum, onesum)
# 出現onesum<0的情況,就設為0,重新累積和
if onesum < 0:
onesum = 0
return maxsum
alist=[-2,1,-3,4,-1,2,1,-5,4]
print(alist)
print(maxSubArray(alist))