LeetCode13 Can I win
2017-05-22 00:04
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1、题目描述
In the “100 game,” two players take turns adding, to a running total, any integer from 1..10. The player who first causes the running total to reach or exceed 100 wins.
What if we change the game so that players cannot re-use integers?
For example, two players might take turns drawing from a common pool of numbers of 1..15 without replacement until they reach a total >= 100.
Given an integer maxChoosableInteger and another integer desiredTotal, determine if the first player to move can force a win, assuming both players play optimally.
You can always assume that maxChoosableInteger will not be larger than 20 and desiredTotal will not be larger than 300.
Example
Input:
maxChoosableInteger = 10
desiredTotal = 11
Output:
false
Explanation:
No matter which integer the first player choose, the first player will lose.
The first player can choose an integer from 1 up to 10.
If the first player choose 1, the second player can only choose integers from 2 up to 10.
The second player will win by choosing 10 and get a total = 11, which is >= desiredTotal.
Same with other integers chosen by the first player, the second player will always win.
2、解题思路
题意:两个玩家从1-maxChoosableInteger之间轮流取值加在一起得到sum,不能重复取值,预先设置一个目标值,desiredTotal,那个玩家最后取完值,此时sum大于等于目标值,就是这个玩家赢,玩家一赢,返回true,否则返回false。
首先直接判断谁赢的情况:目标值小于最大值时可定玩家一赢,返回true,目标值大于所有数字之和时,肯定无解,返回false
其余情况动态规划的思想判断,利用一个数组存储哪些元素已经被选了,然后接下来的游戏便不能选那些已选的元素,接下来的处理流程其实也是一样的问题。当前获胜情况有2种:当前所选取的元素大于还剩余的差;选取当前元素之后,接下来的不能获胜,即对方怎么都输,则自己肯定赢。
3、实现代码
In the “100 game,” two players take turns adding, to a running total, any integer from 1..10. The player who first causes the running total to reach or exceed 100 wins.
What if we change the game so that players cannot re-use integers?
For example, two players might take turns drawing from a common pool of numbers of 1..15 without replacement until they reach a total >= 100.
Given an integer maxChoosableInteger and another integer desiredTotal, determine if the first player to move can force a win, assuming both players play optimally.
You can always assume that maxChoosableInteger will not be larger than 20 and desiredTotal will not be larger than 300.
Example
Input:
maxChoosableInteger = 10
desiredTotal = 11
Output:
false
Explanation:
No matter which integer the first player choose, the first player will lose.
The first player can choose an integer from 1 up to 10.
If the first player choose 1, the second player can only choose integers from 2 up to 10.
The second player will win by choosing 10 and get a total = 11, which is >= desiredTotal.
Same with other integers chosen by the first player, the second player will always win.
2、解题思路
题意:两个玩家从1-maxChoosableInteger之间轮流取值加在一起得到sum,不能重复取值,预先设置一个目标值,desiredTotal,那个玩家最后取完值,此时sum大于等于目标值,就是这个玩家赢,玩家一赢,返回true,否则返回false。
首先直接判断谁赢的情况:目标值小于最大值时可定玩家一赢,返回true,目标值大于所有数字之和时,肯定无解,返回false
其余情况动态规划的思想判断,利用一个数组存储哪些元素已经被选了,然后接下来的游戏便不能选那些已选的元素,接下来的处理流程其实也是一样的问题。当前获胜情况有2种:当前所选取的元素大于还剩余的差;选取当前元素之后,接下来的不能获胜,即对方怎么都输,则自己肯定赢。
3、实现代码
class Solution { public: bool canIWin(int maxChoosableInteger, int desiredTotal) { if (maxChoosableInteger >= desiredTotal) return true; if (maxChoosableInteger * (maxChoosableInteger + 1) / 2 < desiredTotal) return false; unordered_map<int, bool> map; return cIw(maxChoosableInteger, desiredTotal, 0, map); } bool cIw(int length, int total, int used, unordered_map<int, bool>& map) { if (map.count(used)) return map[used]; for (int i = 0; i < length; ++i) { int cur = (1 << i); if ((cur & used) == 0) { if (total <= i + 1 || !cIw(length, total - (i + 1), cur | used, map)) { map[used] = true; return true; } } } map[used] = false; return false; } };
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