zoukankan      html  css  js  c++  java
  • 375. Guess Number Higher or Lower II

    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:

    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.

    Approach #1: DP. [C++]

    class Solution {
    public:
        int getMoneyAmount(int n) {
            vector<vector<int>> dp(n+1, vector<int>(n+1, 0));
            return getMoneyAmount(dp, 1, n);
        }
        
    private:
        int getMoneyAmount(vector<vector<int>>& dp, int s, int e) {
            if (s >= e) return 0;
            if (dp[s][e] != 0) return dp[s][e];
            
            int res = INT_MAX;
            for (int i = s; i <= e; ++i) {
                int tmp = i + max(getMoneyAmount(dp, s, i-1), getMoneyAmount(dp, i+1, e));
                res = min(res, tmp);
            }
            
            dp[s][e] = res;
            
            return res;
        }
    };
    

    Analysis:

    dp[i][j] : the index from i to j how much money you have to pay.

    res = i + max(fuc(), fuc());

    the reasion why we use max instead of min is that we want to find the minimum money which we have to pay. If we use min in this place it will return 0 alway.

    Approach #2: bottom to up. [C++]

        int getMoneyAmount(int n) {
            vector<vector<int>> dp(n+1, vector<int>(n+1, 0));
            for (int i = 2; i <= n; ++i) {
                for (int j = i-1; j > 0; --j) {
                    int globalMin = INT_MAX;
                    for (int k = j+1; k < i; ++k) {
                        int localMin = k + max(dp[j][k-1], dp[k+1][i]);
                        globalMin = min(globalMin, localMin);
                    }
                    dp[j][i] = j+1 == i ? j : globalMin;
                }
            }
            return dp[1][n];
        }
    

      

    Analysis:

    If j+1 == i, dp[j][i] represent two adjacent numbers, so we return the little one as the value of dp[j][i]. why not is the bigger one? I think it is realted to we solve this problem using bottom to up method. so we should get the little one's value firstly.

    永远渴望,大智若愚(stay hungry, stay foolish)
  • 相关阅读:
    转 oracle catalog 库常用脚本
    转 【ORACLE】ORA-12537 问题整理
    转 Trace a specific ORA- error
    15%
    MySQL 存储过程
    MySQL 命令行客户机的分隔符
    MySQL 连接join
    MySQL 正则表达式
    MySQL 日期时间函数
    Arthas 快速入门
  • 原文地址:https://www.cnblogs.com/h-hkai/p/10393272.html
Copyright © 2011-2022 走看看