【LeetCode】486. Predict the Winner 解题报告(Python)
【LeetCode】486. Predict the Winner 解题报告(Python)
标签(空格分隔): LeetCode
作者: 负雪明烛
id: fuxuemingzhu
个人博客: http://fuxuemingzhu.cn/
题目地址:https://leetcode.com/problems/predict-the-winner/description/
题目描述:
Given an array of scores that are non-negative integers. Player 1 picks one of the numbers from either end of the array followed by the player 2 and then player 1 and so on. Each time a player picks a number, that number will not be available for the next player. This continues until all the scores have been chosen. The player with the maximum score wins.
Given an array of scores, predict whether player 1 is the winner. You can assume each player plays to maximize his score.
Example 1:
Input: [1, 5, 2]
Output: False
Explanation: Initially, player 1 can choose between 1 and 2.
If he chooses 2 (or 1), then player 2 can choose from 1 (or 2) and 5. If player 2 chooses 5, then player 1 will be left with 1 (or 2).
So, final score of player 1 is 1 + 2 = 3, and player 2 is 5.
Hence, player 1 will never be the winner and you need to return False.
Example 2:
Input: [1, 5, 233, 7]
Output: True
Explanation: Player 1 first chooses 1. Then player 2 have to choose between 5 and 7. No matter which number player 2 choose, player 1 can choose 233.
Finally, player 1 has more score (234) than player 2 (12), so you need to return True representing player1 can win.
Note:
- 1 <= length of the array <= 20.
- Any scores in the given array are non-negative integers and will not exceed 10,000,000.
- If the scores of both players are equal, then player 1 is still the winner.
题目大意
眼神越来越不好了,看了很多遍题目都看成了预测冬天-_-||
题目的意思很简单,两个玩家可以轮流的从一个数组前面和后面拿数字,每次只能拿一个,被拿走的不能再拿了,最后判断先拿的那个玩家能不能赢。
解题方法
这个题要感谢左程云的讲解,使用两个函数代表先发和后发的情况。如果先发,那么应该返回的是拿走前面的数字或者拿走后面的数字能拿到的结果的最大值。但是如果后发,那么应该返回的是前面不拿和后面不拿的最小值。
直接暴力递归会超时,使用记忆化搜索就能通过了。因为使用的python2,除法默认是地板除,这就是我第一次提交失败的原因,这种情况使用浮点数除法就没问题了。
题目和877. Stone Game做法一样的。
官方解答写了各种方法的对比,包括DP,值得一看:https://leetcode.com/articles/predict-the-winner/
代码如下:
class Solution(object):
def PredictTheWinner(self, nums):
"""
:type nums: List[int]
:rtype: bool
"""
_sum = sum(nums)
self.f_map, self.s_map = dict(), dict()
player1 = self.f(nums, 0, len(nums)-1, self.f_map)
print(player1)
return player1 >= _sum / 2.0
def f(self, nums, start, end, f_map):
if start == end:
return nums[start]
if (start, end) not in f_map:
f_val = max(nums[start] + self.s(nums, start+1, end, self.s_map), nums[end] + self.s(nums, start, end-1, self.s_map))
f_map[(start, end)] = f_val
return f_map[(start, end)]
def s(self, nums, start, end, s_map):
if start == end:
return 0
if (start, end) not in s_map:
s_val = min(self.f(nums, start+1, end, self.f_map), self.f(nums, start, end-1, self.f_map))
s_map[(start, end)] = s_val
return s_map[(start, end)]
上面的代码写复杂了,可以简化成下面这样:
class Solution(object):
def PredictTheWinner(self, nums):
"""
:type nums: List[int]
:rtype: bool
"""
_sum = sum(nums)
self.f_map, self.s_map = dict(), dict()
player1 = self.f(nums, 0, len(nums)-1)
return player1 >= _sum / 2.0
def f(self, nums, start, end):
if start == end:
return nums[start]
if (start, end) not in self.f_map:
f_val = max(nums[start] + self.s(nums, start+1, end), nums[end] + self.s(nums, start, end-1))
self.f_map[(start, end)] = f_val
return self.f_map[(start, end)]
def s(self, nums, start, end):
if start == end:
return 0
if (start, end) not in self.s_map:
s_val = min(self.f(nums, start+1, end), self.f(nums, start, end-1))
self.s_map[(start, end)] = s_val
return self.s_map[(start, end)]
方法二:
DP,未完待续。
参考资料:
日期
2018 年 9 月 8 日 ———— 美好的周末,从刷题开始
【LeetCode】486. Predict the Winner 解题报告(Python)的更多相关文章
- LN : leetcode 486 Predict the Winner
lc 486 Predict the Winner 486 Predict the Winner Given an array of scores that are non-negative inte ...
- [LeetCode] 486. Predict the Winner 预测赢家
Given an array of scores that are non-negative integers. Player 1 picks one of the numbers from eith ...
- 【LeetCode】62. Unique Paths 解题报告(Python & C++)
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址:https://leetcode.com/problems/unique-pa ...
- 随手练——博弈论入门 leetcode - 486. Predict the Winner
题目链接:https://leetcode.com/problems/predict-the-winner/ 1.暴力递归 当前数组左边界:i,右边界:j: 对于先发者来说,他能取到的最大值是:max ...
- [leetcode] 486. Predict the Winner (medium)
原题 思路: 解法一: 转换比较拿取分数多少的思路,改为考虑 player拿的分数为正,把Player2拿的视为负,加上所有分数,如果最后结果大于0则Player1赢. 思考得出递归表达式: max( ...
- 【LeetCode】649. Dota2 Senate 解题报告(Python)
[LeetCode]649. Dota2 Senate 解题报告(Python) 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地 ...
- 【LeetCode】376. Wiggle Subsequence 解题报告(Python)
[LeetCode]376. Wiggle Subsequence 解题报告(Python) 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.c ...
- 【LeetCode】911. Online Election 解题报告(Python)
[LeetCode]911. Online Election 解题报告(Python) 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ ...
- 【LeetCode】886. Possible Bipartition 解题报告(Python)
[LeetCode]886. Possible Bipartition 解题报告(Python) 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu ...
随机推荐
- UE4之Slate:纯C++工程配置
概述: Slate是UE4提供的UI框架,整个UE4 Editor UI都是使用Slate构建的: Slate的官方文档:[Slate UI框架] Slate底层内容,中文环境下能搜索到的有效资源也不 ...
- keras房价预测数据集boston_housing.npz文件下载
github地址: https://github.com/yuxiwang66/boston_housing 码云地址: https://gitee.com/polarisslk/boston_hou ...
- hadoop运行jar包报错
执行命令:[root@hadoop102 mapreduce]# hadoop jar mapreduce2_maven.jar Filter 错误信息:Exception in thread &qu ...
- 我好像发现了一个Go的Bug?
从一次重构说起 这事儿还得从一次重构优化说起. 最近在重构一个路由功能,由于路由比较复杂,需求变化也多,于是想通过责任链模式来重构,刚好这段时间也在 Sentinel-Go 中看到相关源码. 用责任链 ...
- STM32一些特殊引脚做IO使用的注意事项
1 PC13.PC14.PC15的使用 这三个引脚与RTC复用,<STM32参考手册>中这样描述: PC13 PC14 PC15需要将VBAT与VDD连接,实测采用以下程序驱动4个74HC ...
- ORACLE CACHE BUFFER CHAINS原理
原理图如下: 一个cache buffer chains 管理多个hash bucket,受隐含参数:_db_block_hash_buckets(控制管理几个hash bucket)
- Oracle 表结构管理
表其实是数据的'容器'.oracle有几种类型的表: 普通表(ordinary table)又叫堆组织表. 聚簇表(clustered table) 分区表(partition table) 外部表( ...
- grep命令输出显示高亮字
grep命令执行后,终端上输出显示颜色可以加"--color=auto"的参数. 另外的两个办法是: 1.设置环境变量: export GREP_OPTIONS="--c ...
- Android 内存泄漏检测工具 LeakCanary(Kotlin版)的实现原理
LeakCanary 是一个简单方便的内存泄漏检测框架,做 android 的同学基本都收到过 LeakCanary 检测出来的内存泄漏.目前 LeakCanary 最新版本为 2.7 版本,并且采用 ...
- 在vue3中使用router-link-active遇到的坑
在使用 router-link-active 设置链接激活时CSS类名时,发现在例如 /member/order 和 /member/order/:id 这两个都包含 /member/order的路由 ...