动态规划常见问题
零、组合问题
1、硬币问题
n=len(arr)
if n<=0 or aim<=0:
return 0
dp=[float("inf")]*(aim+1)
dp[0]=0
for i in range(n):
for j in range(arr[i],aim+1):
dp[j]=min(dp[j],dp[j-arr[i]]+1)
return dp[aim] if dp[aim]!=float("inf") else -1
https://www.nowcoder.com/practice/3911a20b3f8743058214ceaa099eeb45?tpId=117&&tqId=37795&rp=1&ru=/activity/oj&qru=/ta/job-code-high/question-ranking
一、子序列、字串问题
1、最长连续字串
res = ""
maxlen = 0
for i in range(len(str1)):
if str1[i-maxlen : i+1] in str2:
res = str1[i-maxlen:i+1]
maxlen = maxlen + 1
return res
https://www.nowcoder.com/practice/f33f5adc55f444baa0e0ca87ad8a6aac?tpId=117&&tqId=37799&rp=1&ru=/activity/oj&qru=/ta/job-code-high/question-ranking
2、最长子序列（不要求连续）长度
m, n = len(text1), 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], dp[i][j - 1])
return dp[m][n]
https://leetcode-cn.com/problems/longest-common-subsequence/
3、最长子序列（不要求连续）
m=len(s1)
n=len(s2)
dp=[["" for j in range(n+1)] for i in range(m+1)]
for i in range(m):
for j in range(n):
if s1[i]==s2[j]:
dp[i+1][j+1]=dp[i][j]+s1[i]
else:
if len(dp[i][j+1])>len(dp[i+1][j]):
dp[i+1][j+1]= dp[i][j+1]
else:
dp[i+1][j+1]= dp[i+1][j]
return dp[m][n] if dp[m][n]!="" else "-1"
https://www.nowcoder.com/practice/6d29638c85bb4ffd80c020fe244baf11?tpId=117&&tqId=37798&rp=1&ru=/activity/oj&qru=/ta/job-code-high/question-ranking
二、股票交易问题
二次交易
if len(prices)<=1:
return 0
k=len(prices)
#定义最大收益
dp0=[0]*k
dp1=[0]*k
dp2=[0]*k
dp3=[0]*k
dp4=[0]*k
dp0[0]=0
dp1[0]=-prices[0]
dp2[0]=0
dp3[0]=-prices[0]
dp4[0]=0
for i in range(1,k):
dp0[i]=dp0[i-1]
dp1[i]=max(dp1[i-1],dp0[i-1]-prices[i])
dp2[i]=max(dp2[i-1],dp1[i-1]+prices[i])
dp3[i]=max(dp3[i-1],dp2[i-1]-prices[i])
dp4[i]=max(dp4[i-1],dp3[i-1]+prices[i])
return dp4[k-1]
https://www.nowcoder.com/practice/4892d3ff304a4880b7a89ba01f48daf9?tpId=117&&tqId=37847&rp=1&ru=/activity/oj&qru=/ta/job-code-high/question-ranking
三、子数组最大乘积
核心的关键是：判断当前值的正负以及需要两个状态
k=len(arr)
dpmax=[0]*k
dpmin=[0]*k
if k==0:
return
if k==1:
return arr[0]
dpmax[0]=arr[0]
dpmin[0]=arr[0]
maxsum=arr[0]
for i in range(1,k):
if arr[i]>0:
dpmax[i]=max(arr[i],arr[i]*dpmax[i-1])
dpmin[i]=min(arr[i],arr[i]*dpmin[i-1])
else:
dpmax[i]=max(arr[i],arr[i]*dpmin[i-1])
dpmin[i]=min(arr[i],arr[i]*dpmax[i-1])
maxsum=max(maxsum,dpmax[i])
return maxsum
https://www.nowcoder.com/practice/9c158345c867466293fc413cff570356?tpId=117&&tqId=37785&rp=1&ru=/activity/oj&qru=/ta/job-code-high/question-ranking
四、字符串的全排
res=[]
c=list(ss)
k=len(c)
if k==1:
return c
else:
for i in range(k):
first=str(c[i])
for tmp in self.Permutation(''.join(c[:i]+c[i+1:])):
second=''.join(tmp)
s=first+second
if s not in res:
res.append(s)
return res
https://www.nowcoder.com/practice/fe6b651b66ae47d7acce78ffdd9a96c7?tpId=117&&tqId=37778&rp=1&ru=/activity/oj&qru=/ta/job-code-high/question-ranking
leetcode:567字符串的排列
https://leetcode-cn.com/problems/permutation-in-string/
双指针
剑指offer38:字符串的排列
https://leetcode-cn.com/problems/zi-fu-chuan-de-pai-lie-lcof/
回溯法
https://leetcode-cn.com/problems/zi-fu-chuan-de-pai-lie-lcof/solution/mian-shi-ti-38-zi-fu-chuan-de-pai-lie-hui-su-fa-by/
https://leetcode-cn.com/problems/zi-fu-chuan-de-pai-lie-lcof/solution/mian-shi-ti-38-zi-fu-chuan-de-pai-lie-hui-su-fa-by/
