最长公共子序列（LCS）和最长公共连续子串

import java.util.ArrayList;
import java.util.List;

/**
* Created by gaoji on 2017/4/12.
*/
public class LeetcodeSolution {

public static void main(String[] args) {
System.out.println(longestSub("bdcaba","abcbdab"));
longestConsectiveString("abcgooglecbaleABVC","cbagoogleABVC");
}

public static int longestSub(String s1,String s2){
int m = s1.length();
int n = s2.length();
int[][] dp = new int[m+1][n+1];
for (int i = 0; i <= m; i++) {
dp[i][0] = 0;
}
for (int i = 1; i < n+1 ; i++) {
dp[0][i] = 0;
}
for (int i = 1; i < m+1; i++) {
for (int j = 1; j <= n; j++) {
if(s1.charAt(i-1)==s2.charAt(j-1)){
dp[i][j] = dp[i-1][j-1]+1;
}else{
dp[i][j] = Math.max(dp[i][j-1],dp[i-1][j]);
}
}
}

return dp[m]
;
}
//输出两个字符串的最长连续公共子串，如果有多个，都输出。
public static void longestConsectiveString(String s1,String s2){
int m = s1.length();
int n= s2.length();
List<Integer> list = new ArrayList<Integer>();
int[][] dp = new int[m]
;
int length = 0;
int index = -1;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if(i==0||j==0){
if(s1.charAt(i)==s2.charAt(j)){
dp[i][j]=1;
}else{
dp[i][j]=0;
}
}else if(s1.charAt(i)==s2.charAt(j)){
dp[i][j]= dp[i-1][j-1]+1;
if(length<dp[i][j]){
list.clear();
length = dp[i][j];
list.add(i);
}else if(length == dp[i][j]){
list.add(i);
}
}
}
}
for (int i :list) {
System.out.println(s1.substring(i- length + 1, i + 1));
}
}
}

LCS状态转移方程：

       - 如果s1[m] == s2
，则dp[m,n] = dp[m - 1, n -1] + 1；

       - 如果s1[m] != s2
，则dp[m,n] = max{dp[m,n - 1]，dp[m - 1, n]}

最长连续公共子串的状态转移方程：

s1[i]==s2[j],dp[i][j] = dp[i-1][j-1]+1;

s1[i]!=s2[j].dp[i][j] = 0;