题解 | #最长上升子序列(一)# 【动态规划】/【贪心+二分】

方法一：动态规划

• 时间复杂度：$O(n^2)O(n^2)$
• 空间复杂度：$O(n)O(n)$

import java.io.*;
import java.util.*;

public class Main {
    public static void main(String[] args) throws IOException {
        StreamTokenizer st = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        PrintWriter pw = new PrintWriter(System.out);

        st.nextToken();
        int n = (int) st.nval;

        int[] nums = new int[n];
        for (int i = 0; i < n; i++) {
            st.nextToken();
            nums[i] = (int) st.nval;
        }
        
        // f[i] 表示以下标为i的数字结尾的最长上升子序列的长度
        int[] f = new int[n];
        for (int i = 0; i < n; i++) {
            f[i] = 1;
            for (int j = 0; j < i; j++) {
                if (nums[j] < nums[i]) {
                    f[i] = Math.max(f[i], f[j] + 1);
                }
            }
        }
        
        int ans = 1;
        for (int i = 0; i < n; i++) {
            ans = Math.max(ans, f[i]);
        }
        pw.println(ans);

        pw.flush();
        pw.close();
    }
}

贪心 + 二分

• 时间复杂度：$O(n\log n)O(n\; \backslash log\; n)$
• 空间复杂度：$O(n)O(n)$

import java.io.*;
import java.util.*;

public class Main {
    public static void main(String[] args) throws IOException {
        StreamTokenizer st = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        PrintWriter pw = new PrintWriter(System.out);

        st.nextToken();
        int n = (int) st.nval;

        int[] nums = new int[n];
        for (int i = 0; i < n; i++) {
            st.nextToken();
            nums[i] = (int) st.nval;
        }
        
        // q[i] 表示长度为i的上升子序列的结尾数字
        int[] q = new int[n + 1];
        
        int ans = 0;
        for (int i = 0; i < n; i++) {
            // 二分找到小于nums[i]的最大的那个
            int l = 0, r = ans;
            while (l < r) {
                int mid = l + r + 1 >> 1;
                if (q[mid] < nums[i]) {
                    l = mid;
                } else {
                    r = mid - 1;
                }
            }
            q[r + 1] = nums[i];
            ans = Math.max(ans, r + 1);
        }
        pw.println(ans);

        pw.flush();
        pw.close();
    }
}