Leetcode 464. Can I Win 玩游戏 解题报告

1 解体思想

题目前面是介绍了两个人玩100游戏。。后面出了个变种，我就不普及原始规则了，直接说新规则好了

两个人玩游戏，给了两个数maxChoosableInteger ，desiredTotal，从第一个人开始，可以从1到maxChoosableInteger 当中选一个数字（不能用用过的，第一个人第一轮可以随便选），加到当前的累计和当中（这个从0开始），若此时超过desiredTotal，那么第一个人赢，不然换第二个人，每个人开始前都不能选前面已经用过的数字，第二个人选一个数同样累加，若超过desiredTotal第二个人赢，不然交给第一个人，重复这个步骤

现在问说，有没有什么策略可以保证第一个一定赢？

我以为有什么数学方式，结果试了发现没有。。最后看了discuss还是只能dfs暴力解题，顺带加个map提速（因为可能会有重复状态）

代码基本就是照着discuss改的了，原来600ms ac的，现在127ms，不知道后面还能AC么

总觉得是不是有什么数学技巧？求指教

2 原题

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.

3 AC解

public class Solution {
public boolean canIWin(int maxChoosableInteger, int desiredTotal) {
if (desiredTotal<=0) return true;
//如果1到最大能选的值所有和都不能满足目标值，那么肯定失败
if (maxChoosableInteger*(maxChoosableInteger+1)/2<desiredTotal) return false;
char state[] = new char[maxChoosableInteger];
for(int i=0;i<maxChoosableInteger;i++) state[i] = '0';
return dfs(desiredTotal, state, new HashMap<>());
}
private boolean dfs(int total, char[] state, HashMap<String, Boolean> hashMap) {
String key= new String(state);
if (hashMap.containsKey(key)) return hashMap.get(key);
for (int i=0;i<state.length;i++) {
if (state[i]=='0') {
state[i]='1';
if (total<=i+1 || !dfs(total-(i+1), state, hashMap)) {
hashMap.put(key, true);
state[i]='0';
return true;
}
state[i]='0';
}
}
hashMap.put(key, false);
return false;
}

}