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.
这道题给了一个小游戏,有一个数组,两个玩家轮流取数,说明了只能从开头或结尾取,问我们第一个玩家能赢吗。这道题我想到了应该是用Minimax来做,由于之前有过一道这样的题Guess Number Higher or Lower II,所以依稀记得应该要用递归的方法,而且当前玩家赢返回true的条件就是递归调用下一个玩家输返回false。我们需要一个变量来标记当前是第几个玩家,还需要两个变量来分别记录两个玩家的当前数字和,在递归函数里面,如果当前数组为空了,我们直接比较两个玩家的当前得分即可,如果数组中只有一个数字了,我们根据玩家标识来将这个数字加给某个玩家并进行比较总得分。如果数组有多个数字,我们分别生成两个新数组,一个是去掉首元素,一个是去掉尾元素,然后根据玩家标识分别调用不同的递归,只要下一个玩家两种情况中任意一种返回false了,那么当前玩家就可以赢了,参见代码如下:
class Solution {
public:
bool PredictTheWinner(vector<int>& nums) {
int n = nums.size();
vector<vector<int>> dp(n, vector<int>(n, -1));
return canWin(nums, 0, n - 1, dp) >= 0;
}
int canWin(vector<int>& nums, int s, int e, vector<vector<int>>& dp) {
if (dp[s][e] == -1) {
dp[s][e] = (s == e) ? nums[s] : max(nums[s] - canWin(nums, s + 1, e, dp), nums[e] - canWin(nums, s, e - 1, dp));
}
return dp[s][e];
}
};