53.最大子数组和
#dpi代表以nums[i]为结尾的最大子数组和,if dpi>0 then dpi+1=dpi+nums[i+1] else dpi=nums[i+1]
#res=max([dp0,dp1,...,dpn])
dp=nums[0]
ans=nums[0]
for i in range(1,len(nums)):
if dp>0:
dp+=nums[i]
else:
dp=nums[i]
ans=max(ans,dp)
return ans
121. 买卖股票的最佳时机
#1.卖出时,只要在卖出前价格最低的一天买入,即利润最大化
minPrice=prices[0]
maxProfit=0
for i in range(1,len(prices)):
minPrice=min(minPrice,prices[i])
maxProfit=max(maxProfit,prices[i]-minPrice)
return maxProfit
#2.dp标准格式
if not prices:
return 0
minPrice=prices[0]
n=len(prices)
dp=[0]*n
for i in range(1,n):
minPrice=min(prices[i],minPrice)
dp[i]=max(dp[i-1],prices[i]-minPrice)
return dp[-1]
#3.dp[i][0]表示第i天不持股的最大利润、dp[i][1]表示第i天持股的最大利润
if not prices:
return 0
n=len(prices)
dp=[[0]*2 for _ in range(n)]
dp[0][0]=0
dp[0][1]=-prices[0]
for i in range(1,n):
dp[i][0]=max(dp[i-1][0],dp[i-1][1]+prices[i])
dp[i][1]=max(dp[i-1][1],-prices[i])
return dp[-1][0]
5. 最长回文子串
#dp[i][j]表示索引为i-j的子串是否回文,dp[i][j]与dp[i+1][j-1]有关,因此for循环先遍历i后遍历j。
if not s:
return ""
n = len(s)
dp = [[0] * n for _ in range(n)]
maxLen = 1
idx = 0
for i in range(n):
dp[i][i] = 1
for j in range(1, n):
for i in range(0, j):
if s[i] == s[j] and (dp[i + 1][j - 1] == 1 or j == i + 1):
dp[i][j] = 1
if j - i + 1 > maxLen:
maxLen = j - i + 1
idx = i
return s[idx:idx + maxLen]
下面两题注意区别,公共子序列和重复子数组,细微差别!!!!
1143. 最长公共子序列
class Solution:
def longestCommonSubsequence(self, text1: str, text2: str) -> int:
m = len(text1)
n = len(text2)
dp=[[0]*(n+1) for _ in range(m+1)]
for i in range(1,m+1):
for j in range(1,n+1):
if text1[i-1]==text2[j-1]:
dp[i][j]=dp[i-1][j-1]+1
else:
dp[i][j]=max(dp[i-1][j-1],dp[i-1][j],dp[i][j-1])
return dp[-1][-1]
718. 最长重复子数组
class Solution:
def findLength(self, nums1: List[int], nums2: List[int]) -> int:
# dp
res=0
if not nums1 or not nums2:
return res
m, n = len(nums1), len(nums2)
dp = [[0]*(n+1) for _ in range(m+1)]
for j in range(1, n+1):
for i in range(1, m+1):
'''A 、B数组各抽出一个前缀数组,单看它们的末尾项,如果它们俩不一样,则公共子数组肯定不包括它们俩。
如果它们俩一样,则要考虑它们俩前面的子数组「能为它们俩提供多大的公共长度」'''
if nums1[i-1] == nums2[j-1]:
dp[i][j] = dp[i-1][j-1]+1
else:
dp[i][j]=0
if dp[i][j]>res:
res=dp[i][j]
return res