51Nod 1006 最长公共子序列Lcs

动态规划，转移方程为：字符相同dp[i][j]=dp[i-1][j-1]+1；不同dp[i][j]=MAX(dp[i-1][j],dp[i][j-1])

最后根据MAX值和dp[i][j]（dp[i][j]始终是最大值）来判断哪个字符相同，则存进stringbuilder中。

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class LCS {
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        String s1 = in.readLine();
        String s2 = in.readLine();

        char[] str1 = s1.toCharArray();
        char[] str2 = s2.toCharArray();

        int len1 = str1.length;
        int len2 = str2.length;
        int[][] dp = new int[len1+1][len2+1];
        //初始化
        dp[0][0] = 0;
        for (int i = 0; i <=len1; i++) {
            dp[i][0] = 0;
        }
        for (int i = 0; i <=len2; i++) {
            dp[0][i] = 0;
        }
        //计算DP值
        for (int i = 1; i <=len1; i++) {
            for (int j = 1; j <=len2; j++) {
                if (str1[i-1] == str2[j-1])
                    dp[i][j] = dp[i-1][j-1]+1;
                else
                    dp[i][j]= Math.max(dp[i-1][j], dp[i][j-1]);
            }
        }
        StringBuilder stringBuilder = new StringBuilder();//因为String经常更改会导致性能低下，StringBuilder经常改动效率最高
        int k = 0,i = len1 , j = len2;
        //System.out.println(dp[len1][len2]);
        while (k != dp[len1][len2]){
            if (dp[i-1][j] == dp[i][j])//当前字符与另一字符串相对应字符不同
                i--;
            else if (dp[i][j-1] == dp[i][j])//当前字符与另一字符串相对应字符不同
                j--;
            else {
                if (dp[i][j] != dp[i-1][j-1]) {//当前字符与另一字符串相对应字符相同
                    stringBuilder.insert(0, str1[i - 1]);//存进去
                    k++;
                }
                i--;
                j--;
            }
        }
        System.out.println(stringBuilder.toString());
    }
}