AcWing 1014. 登山JAVA(最长单调子序列 单调栈)

时间复杂度：

O(nlogn)

//单调栈问题
//分别计算从左向右和从右向左的最长上升子序列,分别存储到两个数组中
import java.util.*;
 
class Main{
    static int n = 0, N = 1010;
    static int[] nums = new int[N];
    static int[] left = new int[N], right = new int[N];
    static int bs(int[] max, int len, int num){//找到从左到右第一个大于num的数字
        int l = 0, r = len;
        while(l < r){
            int mid = l + r >> 1;//+1防止mid不变造成死循环。
            if(max[mid] < num)l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static void main(String[] args)throws Exception{
        Scanner sc = new Scanner(System.in);
        int m = sc.nextInt();
        for(int i = 1; i <= m; ++i)nums[i] = sc.nextInt();
        int[] st = new int[N];//单调栈
        int tt = -1;
        for(int i = 1; i <= m; ++i){
            if(tt == -1)st[++tt] = nums[i];
            if(tt >= 0 && st[tt] > nums[i]){
                int index = bs(st, tt, nums[i]);
                st[index] = nums[i];
            }
            if(st[tt] < nums[i])st[++tt] = nums[i];
            left[i] = tt + 1;
        }
        tt = -1;
        for(int i = m; i >= 1; --i){
            if(tt == -1)st[++tt] = nums[i];
            if(tt >= 0 && st[tt] > nums[i]){
                int index = bs(st, tt, nums[i]);
                st[index] = nums[i];
            }
            if(st[tt] < nums[i])st[++tt] = nums[i];
            right[i] = tt + 1;
        }
        int max = 0;
        for(int i = 1; i <= m; ++i){
            max = Math.max(max, left[i] + right[i] - 1);
        }
        System.out.print(max);
        
    }
}