字符串最长公共子序列问题

Posted 2020-08-09 ~风轻云淡~

找两个字符串的最长公共子序列，最长公共子序列并不要求连续。

代码如下：

package string;

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

/**
 * 字符串的最长公共子序列问题
 * @author Administrator
 *
 */
public class LCSequence {

    /**
     * 求最长公共子序列长度
     * @param s1
     * @param s2
     * @return
     */
    public int getMaxLCSLen(String s1, String s2){
        int maxLen = 0;
        if(s1 == null || s2 == null){
            return maxLen;
        }
        int m = s1.length();
        int n = s2.length();
        // a[i][j]记录s1[0~i-1]与s2[0~j-1]的最长公共子序列长度
        int[][] a = new int[m+1][n+1];
        for(int i = 1; i <= m; i++){
            for(int j = 1; j <= n; j++){
                if(s1.charAt(i-1) == s2.charAt(j-1)){
                    a[i][j] = Math.max(a[i][j-1], a[i-1][j-1] + 1);
                    a[i][j] = Math.max(a[i][j], a[i-1][j]);
                }else{
                    a[i][j] = Math.max(a[i][j-1], a[i-1][j]);
                }
                maxLen = Math.max(maxLen, a[i][j]);
            }
        }
        return maxLen;
    }
    
    /**
     * 求最长公共子序列
     * @param s1
     * @param s2
     * @return
     */
    public List<String> getMaxLCS(String s1, String s2){
        int maxLen = 0;
        List<String> res = new ArrayList<String>();
        if(s1 == null || s2 == null){
            return res;
        }
        int m = s1.length();
        int n = s2.length();
        // a[i][j]记录s1[0~i-1]与s2[0~j-1]的最长公共子序列长度
        int[][] a = new int[m+1][n+1];
        for(int i = 1; i <= m; i++){
            for(int j = 1; j <= n; j++){
                if(s1.charAt(i-1) == s2.charAt(j-1)){
                    a[i][j] = Math.max(a[i][j-1], a[i-1][j-1] + 1);
                    a[i][j] = Math.max(a[i][j], a[i-1][j]);
                }else{
                    a[i][j] = Math.max(a[i][j-1], a[i-1][j]);
                }
                if(a[i][j] == maxLen){
                    String s = s1.substring(i-a[i][j], i);
                    if(!res.contains(s)){
                        res.add(s);
                    }
                } else if(a[i][j] > maxLen){
                    maxLen = a[i][j];
                    res = new ArrayList<String>();
                    res.add(s1.substring(i-a[i][j], i));
                }
            }
        }
        return res;
    }
    
    
    public static void main(String[] args) {
        LCSequence lcs = new LCSequence();
        String s1 = "a1b2c3";
        String s2 = "1a1wbz2c123a1b2c123";
        System.out.println(lcs.getMaxLCSLen(s1, s2));
        System.out.println(lcs.getMaxLCS(s1, s2));
    }

}