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 particula…

[LeetCode]375. Guess Number Higher or Lower II 解题报告(Python) 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址: https://leetcode.com/problems/guess-number-higher-or-lower-ii/description/ 题目描述: We are playing the Guess Game. The game is as f…

好久没切 leetcode 的题了,静下心来切了道,这道题比较有意思,和大家分享下. 我把它叫做 "不一样的猜数字游戏",我们先来看看传统的猜数字游戏,Guess Number Higher or Lower.题意非常的简单,给定一个数字 n,系统会随机从 1 到 n 中抽取一个数字,你需要写一个函数 guessNumber,它的作用是返回系统选择的数字,同时你还有一个外部的 API 可以调用,是为 guess 函数,它会将你猜的数字和系统选择的数字比较,是大了还是小了. 非常的简单,…

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 particula…

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 particula…

374. Guess Number Higher or Lower 二分查找就好 // Forward declaration of guess API. // @param num, your guess // @return -1 if my number is lower, 1 if my number is higher, otherwise return 0 int guess(int num); class Solution { public: int guessNumber(int…

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 particula…

最后更新 四刷? 极大极小算法..还是叫极小极大的.. 首先要看怎么能保证赢. 比如2个数,猜第一个猜第二个都能保证下一轮我们赢定了,为了少交钱,我们猜小的. 比如3个数,猜第二个才能保证下一轮再猜一定能赢. 比如4个数,怎么猜都不能保证下一轮一定赢,所以要猜两轮.第一轮猜1,2,3还是4就很讲究了. 猜1就是num1 + dp[2][4](今后花费) 猜2就是dp[1][1](今后花费) + num2或者num[1] + dp[3][4](今后花费) ... 4种里每种都算出来,看哪种花费最少…

我们正在玩一个猜数游戏,游戏规则如下:我从 1 到 n 之间选择一个数字,你来猜我选了哪个数字.每次你猜错了,我都会告诉你,我选的数字比你的大了或者小了.然而,当你猜了数字 x 并且猜错了的时候,你需要支付金额为 x 的现金.直到你猜到我选的数字,你才算赢得了这个游戏.示例:n = 10, 我选择了8.第一轮: 你猜我选择的数字是5,我会告诉你,我的数字更大一些,然后你需要支付5块.第二轮: 你猜是7,我告诉你,我的数字更大一些,你支付7块.第三轮: 你猜是9,我告诉你,我的数字更小一些,你支付…

原题 思路: miniMax+DP dp[i][j]保存在i到j范围内,猜中这个数字需要花费的最少 money. "至少需要的花费",就要我们 "做最坏的打算,尽最大的努力",即取最大值. dp[beg][end] =MIN ( i + max(helper(beg, i - 1, dp), helper(i + 1, end, dp)) ) class Solution { public: int helper(int beg, int end, vector&l…