题解 | 最长公共子序列(二)

import java.util.*;


public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * longest common subsequence
     * @param s1 string字符串 the string
     * @param s2 string字符串 the string
     * @return string字符串
     */
    public String LCS (String s1, String s2) {
        int len1 = s1.length();
        int len2 = s2.length();
        if (len1 == 0 || len2 == 0)
            return "-1";
        int[][] dp = new int[len1 + 1][len2 + 1];
        for (int i = 0; i < len1 + 1; i++) {
            for (int j = 0; j < len2 + 1; j++) {
                //初始化行列第一个元素
                if (i == 0 || j == 0) {
                    dp[i][j] = 0;
                    continue;
                }
                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 - 1][j], dp[i][j - 1]);
                }
            }
        }
        //找出一个最长的公共子序列
        StringBuilder sb = new StringBuilder();
        int s1L = len1, s2L = len2;
        while (s1L != 0 && s2L != 0) {
            //如果字符相等,则记录结果，都往后退一格
            if (s1.charAt(s1L - 1) == s2.charAt(s2L - 1)) {
                sb.append(s1.charAt(s1L - 1));
                s1L--;
                s2L--;
            } else {
                //比较上一个值谁大,大的往后退一格
                if (dp[s1L - 1][s2L] > dp[s1L][s2L - 1]) {
                    s1L--;
                } else {
                    s2L--;
                }
            }
        }
        if (sb.length() == 0) {
            return "-1";
        }
        return sb.reverse().toString();
    }
}