动态规划源码

package leetcode;

import java.util.Scanner;

public class DP {

static int N =100;

static int dp[][]=new int[N][N]; //动态规划矩阵

static char result[]=new char[N];

static int index;

//Section 1[求解最长公共子序列]：不需要子序列连续

public static void initDPfromBeginToEnd(int lena,int lenb)

{

int i,j;

for(i=0;i<=lena;i++)

dp[i][0]=0;

for(j=0;j<=lenb;j++)

dp[0][j]=0;

}

//从前往后构造dp矩阵

public static void changeDPfromBeginToEnd(char []s1,int lena,char []s2,int lenb) {

int i,j;

for(i=1;i<=lena;i++)

{

for(j=1;j<=lenb;j++)

{

if(s1[i-1]==s2[j-1]) //从第一个[index=0]字符开始

{

dp[i][j]=dp[i-1][j-1]+1;

}

else

{

dp[i][j]=Math.max(dp[i-1][j],dp[i][j-1]);

}

}

}

}

public static void initDPfromEndToBegin(int lena,int lenb)

{

int i,j;

for(i=0;i<=lena;i++)

dp[i][lenb]=0;

for(j=0;j<=lenb;j++)

dp[lena][j]=0;

}

//从后往前构造dp矩阵

public static void changeDPfromEndToBegin(char []s1,int lena,char []s2,int lenb) {

int i,j;

for(i=lena-1;i>=0;i--)

{

for(j=lenb-1;j>=0;j--)

{

if(s1[i]==s2[j])

{

dp[i][j]=dp[i+1][j+1]+1;

}

else

{

dp[i][j]=Math.max(dp[i+1][j],dp[i][j+1]);

}

}

}

}

public static void printDP(int n,int m)

{

int i,j;

for(i=0;i<=n;i++)

{

for(j=0;j<=m;j++)

{

System.out.print(dp[i][j]+" ");

}

System.out.println();

}

}

//Section 2[求解最长公共子串]：需要子串连续

public static String getLCSLength(String s, String t)

{

int p = s.length();

int q = t.length();

String [][]num = new String [p][];

for(int i=0;i

{

num[i] = new String[q];

}

char char1 = '\0';

char char2 = '\0';

int len = 0;

String lcs = "" ;

for(int i=0; i

{

for (int j=0; j

{

char1 = s.charAt(i);

char2 = t.charAt(j);

if(char1!=char2)

{

num[i][j] = "" ;

}

else

{

if (i==0||j==0)

{

num[i][j]=String.valueOf(char1);

}

else

{

num[i][j]=num[i-1][j-1]+char1;

}

if(num[i][j].length()>len)

{

len = num[i][j].length() ;

lcs = num[i][j];

}

else if(num[i][j].length()==len)

{

lcs = lcs + "," + num[i][j];

}

}

}

}

return lcs;

}

//Section 3[求解字符串的编辑距离]

public static int min3(int x,int y ,int z)

{

int tmp=Math.min(x,y);

return Math.min(tmp,z);

}

//从前往后

public static void initDPToEnd(int lena,int lenb)

{

int i,j;

for(i=0;i<=lena;i++)

dp[i][0]=i;

for(j=0;j<=lenb;j++)

dp[0][j]=j;

}

public static void changeDPToEnd(char []s1,int lena,char []s2,int lenb) {

int i,j;

index=0;

for(i=1;i<=lena;i++)

{

for(j=1;j<=lenb;j++)

{

if(s1[i-1]==s2[j-1]) //从第一个[index=0]字符开始

{

dp[i][j]=dp[i-1][j-1];

}

else

{

dp[i][j]=min3(dp[i-1][j-1]+1,dp[i-1][j]+1,dp[i][j-1]+1);

}

}

}

}

//从后往前

public static void initDPToBegin(int lena,int lenb)

{

int i,j;

for(i=lena;i>=0;i--)

dp[i][lenb]=lena-i;

for(j=lenb;j>=0;j--)

dp[lena][j]=lenb-j;

}

public static void changeDPToBegin(char []s1,int lena,char []s2,int lenb) {

int i,j;

index=0;

for(i=lena-1;i>=0;i--)

{

for(j=lenb-1;j>=0;j--)

{

if(s1[i]==s2[j])

{

dp[i][j]=dp[i+1][j+1];

}

else

{

dp[i][j]=min3(dp[i+1][j+1]+1,dp[i+1][j]+1,dp[i][j+1]+1);

}

}

}

}

//Section 4[求解字符串里面的最长回文子串]

//使用【暴力1】法求解

public static boolean IsPalindrome(char []str,int start ,int end)

{

while(start

{

if(str[start]!=str[end])

break;

else

{

start++;

end--;

}

}

if(start==end)

return true;

else

return false;

}

public static char[]getString(char[]str,int start,int end)

{

char []result=new char[end-start+1]; //不需要设置串尾'0'

int index,i;

index=0;

for(i=start;i<=end;i++)

result[index++]=str[i];

//result[index+1]='\0'; //注意java中没有字符串的串尾的说法return result;

}

//时间复杂度0(n^2)

public static void getLongestPalindromeString(char[]str)

{

int i,j,len,maxlen,start,end;

len=str.length;

maxlen=0;

start=end=0;

for(i=0;i

{

for(j=i+1;j

{

if(IsPalindrome(str,i,j))

{

int currlen=j-i+1;

if(currlen>maxlen)

{

maxlen=currlen;

start=i;

end=j;

}

}

}

}

char[]resultString=getString(str,start,end);

System.out.println(resultString);

}

//使用【暴力2】法求解

//以某一字符为中心[左右对称]的方法

//以abccba为例

public static void getLongestPalindromeString_1(char []str)

{

int i,j,len,maxlen,start,end;

len=str.length;

maxlen=0;

start=end=0;

j=0;

for(i=0;i

{

//实现对称中心化

if(i-j>=0 && i+j

{

if(maxlen<2*j+1)

{

maxlen=2*j+1;

start=i-j;

end=i+j;

}

j++;

}

else

{

i++;

j=0;

}

}

char[]resultString=getString(str,start,end);

System.out.println(resultString);

}

///使用【动态规划3】法求解

public static void getLongestPalindromeString_2(char[]str) {

int len=str.length;

for (int i=len-1;i>=0;i--)

{

dp[i][i]=1;

for (int j=i+1;j

if (str[i]==str[j])

dp[i][j]=dp[i+1][j-1]+2;

else

dp[i][j]=Math.max(dp[i][j-1],dp[i+1][j]);

}

System.out.println(dp[0][len-1]);

}

public static void main(String []args)

{

char s1[]="abcd".toCharArray();

char s2[]="cxbydz".toCharArray();

int lena,lenb;

lena=s1.length;

lenb=s2.length;

//1)最长公共子序列

//使用从前往后的构造法initDPfromBeginToEnd(lena,lenb); changeDPfromBeginToEnd(s1,lena,s2,lenb); printDP(lena,lenb);

System.out.println(dp[lena][lenb]);

System.out.println();

//使用从后往前的构造法initDPfromEndToBegin(lena,lenb); changeDPfromEndToBegin(s1,lena,s2,lenb); printDP(lena,lenb);

System.out.println(dp[0][0]);

System.out.println();

//3)字符串的编辑距离

//使用从前往后的构造法

initDPToEnd(lena,lenb);

changeDPToEnd(s1,lena,s2,lenb);

printDP(lena,lenb);

System.out.println(dp[lena][lenb]);

System.out.println();

//使用从后往前的构造法

initDPToBegin(lena,lenb);

changeDPToBegin(s1,lena,s2,lenb);

printDP(lena,lenb);

System.out.println(dp[0][0]);

System.out.println();

//4)字符串的最长回文子串

char test[]="abcdcba1".toCharArray();

System.out.println(IsPalindrome(test,0,test.length-1)); System.out.println(getString(test,0,1)); getLongestPalindromeString(test); getLongestPalindromeString_1(test); getLongestPalindromeString_2(test);

//2)最长公共子串

Scanner input=new Scanner(System.in);

String s,t;

System.out.println("请输入字符串s:");

s=input.next();

System.out.println("请输入字符串t:");

t=input.next();

System.out.println("最长公共子串为:");

System.out.println(getLCSLength(s,t));

}

}