【重点】【DP】1143.最长公共子序列|516.最长回文子序列

两个求解代码类似的题目，对比记忆！！！

1143.最长公共子序列

法1：DP，考虑空串

java 复制代码

class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
        int m = text1.length() + 1, n = text2.length() + 1; // 加1是在构建二维矩阵时增加空串情况, 简化计算
        int[][] dp = new int[m][n];
        for (int i = 1; i < m; ++i) {
            for (int j = 1; j < n; ++j) {
                if (text1.charAt(i - 1) == text2.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 - 1][n - 1];
    }
}

法2：DP，不考虑空串

代码会啰嗦很多，所以还是法1好哇！！！

java 复制代码

class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
        int m = text1.length(), n = text2.length();
        int[][] dp = new int[m][n];
        boolean existed = false;
        for (int i = 0; i < n; ++i) {
            dp[0][i] = (existed || text1.charAt(0) == text2.charAt(i)) ? 1 : 0;
            if (dp[0][i] == 1) {
                existed = true;
            }
        }
        existed = false;
        for (int i = 0; i < m; ++i) {
            dp[i][0] = (existed || text1.charAt(i) == text2.charAt(0)) ? 1 : 0;
            if (dp[i][0] == 1) {
                existed = true;
            }
        }

        for (int i = 1; i < m; ++i) {
            for (int j = 1; j < n; ++j) {
                if (text1.charAt(i) == text2.charAt(j)) {
                    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 - 1][n - 1];
    }
}

516.最长回文子序列

题目

java 复制代码

class Solution {
    public int longestPalindromeSubseq(String s) {
        int n = s.length();
        int[][] dp = new int[n][n];
        for (int i = 0; i < n; ++i) {
            dp[i][i] = 1;
        }
        for (int i = n - 1; i >= 0; --i) {
            for (int j = i + 1; j < n; ++j) {
                if (s.charAt(i) == s.charAt(j)) {
                    dp[i][j] = (j - i == 1 ? 0 : dp[i + 1][j - 1]) + 2;
                } else {
                    dp[i][j] = Math.max(dp[i + 1][j], dp[i][j - 1]);
                }
            }
        }

        return dp[0][n - 1];
    }
}