例题列表

898. 数字三角形

每个数字可以从它上面的左右两边选择一个大的转移过来。

如果想不清楚的话，可以不要节省空间，就写个完整的 n * n 大小的 dp 数组就好。

import java.util.*;

public class Main {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[][] dp = new int[n][n];
        dp[0][0] = sc.nextInt();
        for (int i = 1; i < n; ++i) {
            for (int j = 0; j <= i; ++j) {
                dp[i][j] = sc.nextInt();
                if (j == 0) dp[i][j] += dp[i - 1][j];
                else if (j == i) dp[i][j] += dp[i - 1][j - 1];
                else dp[i][j] += Math.max(dp[i - 1][j - 1], dp[i - 1][j]);
            }
        }
        System.out.println(Arrays.stream(dp[n - 1]).max().getAsInt());
    }
}

895. 最长上升子序列（n^2两重循环dp）

import java.util.*;

public class Main {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] a = new int[n];
        int[] dp = new int[n];
        Arrays.fill(dp, 1);
        for (int i = 0; i < n; ++i) {
            a[i] = sc.nextInt();
            for (int j = 0; j < i; ++j) {
                if (a[i] > a[j]) dp[i] = Math.max(dp[i], dp[j] + 1);
            }
        }
        System.out.println(Arrays.stream(dp).max().getAsInt());
    }
}

896. 最长上升子序列 II（贪心+二分查找）

https://www.acwing.com/problem/content/898/

import java.util.*;

public class Main {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        List<Integer> ls = new ArrayList<>();
        for (int i = 0; i < n; ++i) {
            int a = sc.nextInt();
            if (ls.isEmpty() || a > ls.get(ls.size() - 1)) ls.add(a);
            else {
                // 找到列表第一个>=当前元素的值
                int l = 0, r = ls.size() - 1;
                while (l < r) {
                    int mid = l + r >> 1;
                    if (ls.get(mid) < a) l = mid + 1;
                    else r = mid;
                }
                ls.set(l, a);
            }
        }
        System.out.println(ls.size());
    }
}

897. 最长公共子序列

https://www.acwing.com/problem/content/899/

import java.util.*;

public class Main {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt(), m = sc.nextInt();
        char[] a = sc.next().toCharArray(), b = sc.next().toCharArray();

        int[][] dp = new int[n + 1][m + 1];
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= m; ++j) {
                if (a[i - 1] == b[j - 1]) {
                    dp[i][j] = Math.max(1 + dp[i - 1][j - 1], dp[i][j]);
                } else dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
            }
        }
        System.out.println(dp[n][m]);
    }
}

902. 最短编辑距离

https://www.acwing.com/problem/content/904/

三种编辑方式，对应着从 dp[i - 1][j - 1]、dp[i][j - 1]、dp[i - 1][j] 转移过来。

import java.util.*;

public class Main {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        char[] a = sc.next().toCharArray();
        int m = sc.nextInt();
        char[] b = sc.next().toCharArray();

        int[][] dp = new int[n + 1][m + 1];
        for (int i = 0; i <= n; ++i) {
            Arrays.fill(dp[i], Integer.MAX_VALUE);
            dp[i][0] = i;
        }
        for (int j = 0; j <= m; ++j) dp[0][j] = j;
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= m; ++j) {
                if (a[i - 1] == b[j - 1]) {
                    dp[i][j] = Math.min(dp[i - 1][j - 1], dp[i][j]);
                } else dp[i][j] = 1 + Math.min(Math.min(dp[i - 1][j], dp[i][j - 1]), dp[i - 1][j - 1]);
            }
        }
        System.out.println(dp[n][m]);
    }
}

899. 编辑距离⭐⭐⭐⭐⭐

https://www.acwing.com/problem/content/901/

不要被吓到，算一下时间复杂度：1000 * 1000 * 10 * 10 = 10^8

所以就是上一题编辑距离的简单应用。

import java.util.*;

public class Main {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt(), m = sc.nextInt();
        String[] strs = new String[n];
        for (int i = 0; i < n; ++i) {
            strs[i] = sc.next();
        }
        for (int i = 0; i < m; ++i) {
            String s = sc.next();
            int k = sc.nextInt(), res = 0;
            for (int j = 0; j < n; ++j) {
                if (edit(strs[j], s) <= k) ++res;
            }
            System.out.println(res);
        }
    }

    static int edit(String a, String b) {
        int m = a.length(), n = b.length();
        int[][] dp = new int[m + 1][n + 1];
        for (int i = 0; i <= m; ++i) dp[i][0] = i;
        for (int j = 0; j <= n; ++j) dp[0][j] = j;
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (a.charAt(i - 1) != b.charAt(j - 1)) dp[i][j] = Math.min(dp[i - 1][j - 1], Math.min(dp[i - 1][j], dp[i][j - 1])) + 1;
                else dp[i][j] = dp[i - 1][j - 1];
            }
        }
        return dp[m][n];
    }
}