Max Points on a Line || Guess Number Higher or Lower II

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Max Points on a Line

Question

Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.

Analysis

九章算法

  • map.put((double)Integer.MIN_VALUE,1); 可能整个序列都是相同点，他们的斜率用MIN_VALUE表示
  • 利用dup变量记录有多少个重复的点
  • 每次进入一个新的坐标点比较的时候记住map.clear
  • 计算key的时候必须+0.0，否则无法通过所有点纵坐标相同的情况

Code

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/**
 * Definition for a point.
 * class Point {
 *     int x;
 *     int y;
 *     Point() { x = 0; y = 0; }
 *     Point(int a, int b) { x = a; y = b; }
 * }
 */
public class Solution {
    public int maxPoints(Point[] points) {
        if(points.length<2)
            return points.length;
        Map<Double,Integer> map=new HashMap<Double,Integer>();
        int max=1;
        for(int i=0;i<points.length;i++){
            map.clear();
            map.put((double)Integer.MAX_VALUE,1);
            int dup=0;
            for(int j=i+1;j<points.length;j++){
                double key=0;
                if(points[i].x==points[j].x&&points[i].y==points[j].y){
                    dup++;
                    continue;
                }
                if(points[i].x==points[j].x)
                    key=(double)Integer.MAX_VALUE;
                else
                    key=0.0+(double)(points[i].y-points[j].y)/(double)(points[i].x-points[j].x);
                map.put(key,map.getOrDefault(key,1)+1);
            }
            for(int tmp:map.values()){
                max=Math.max(dup+tmp,max);
            }
        }
        return max;
    }
}

Palindrome Partitioning II

Question

We are playing the Guess Game. The game is as follows:

I pick a number from 1 to n. You have to guess which number I picked.

Every time you guess wrong, I’ll tell you whether the number I picked is higher or lower.

However, when you guess a particular number x, and you guess wrong, you pay $x. You win the game when you guess the number I picked.

Example:

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n = 10, I pick 8.
First round:  You guess 5, I tell you that it's higher. You pay $5.
Second round: You guess 7, I tell you that it's higher. You pay $7.
Third round:  You guess 9, I tell you that it's lower. You pay $9.
Game over. 8 is the number I picked.
You end up paying $5 + $7 + $9 = $21.

Given a particular n ≥ 1, find out how much money you need to have to guarantee a win.

Analysis

中文参考

LeetCode Discuss

最大最小化

Code

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public class Solution {
    public int getMoneyAmount(int n) {
        int[][] table = new int[n+1][n+1];
        return DP(table, 1, n);
    }
    
    int DP(int[][] t, int s, int e){
        if(s >= e) return 0;
        if(t[s][e] != 0) return t[s][e];
        int res = Integer.MAX_VALUE;
        for(int x=s; x<=e; x++){
            int tmp = x + Math.max(DP(t, s, x-1), DP(t, x+1, e));
            res = Math.min(res, tmp);
        }
        t[s][e] = res;
        return res;
    }
}