动态规划：最长子序列问题

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提到最长子序列问题，一下子就能想到动态规划。

衍生题目：添加最少字符使得原字符串变成回文字符串。例题见：nyoj 37.

直接上代码：

<span style="font-size:24px;">import java.util.*;


public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();

        while (N-- > 0) {
            String str1 = sc.next();
            // String str2=sc.next();
            StringBuffer sb = new StringBuffer(str1);
            String str2 = sb.reverse().toString();
            int length1 = longest(str1, str2);
            System.out.println(str1.length() - length1);
            // System.out.println(length1);
        }
    }

    public static int longest(String str1, String str2) {
        char[] s1 = str1.toCharArray();
        // char[] s2=new StringBuffer(s).reverse().toString().toCharArray();
        char[] s2 = str2.toCharArray();
        int m = s1.length;
        int n = s2.length;
        int[][] dp = new int[m][n];
        dp[0][0] = s1[0] == s2[0] ? 1 : 0;
        for (int i = 1; i < n; i++) {
            dp[0][i] = Math.max(dp[0][i - 1], s1[0] == s2[i] ? 1 : 0);
        }
        for (int i = 1; i < m; i++) {
            dp[i][0] = Math.max(dp[i - 1][0], s1[i] == s2[0] ? 1 : 0);
        }
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (s1[i] == s2[j])
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                else
                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
            }
        }
        return dp[m - 1][n - 1];
    }
}
</span>