Predict the Winner

https://leetcode.com/problems/predict-the-winner/description/

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. 1

   <

   = length of the array

   <

   = 20.

2. Any scores in the given array are non-negative integers and will not exceed 10,000,000.

3. If the scores of both players are equal, then player 1 is still the winner.

Thoughts

给定一串数字，两人依次从首或尾选一个，问有没有策略让先选的一定赢。博弈论，也就是minmax问题，即最大化自己收益 == 最小化对方收益 (最大化对方负收益)。同样对方收益也是如此定义。可以看成自己和自己玩，左右互博，因此从问题定义看就是个递归问题: 让score表示当前玩家最大收益，它等于max_{所有选择}(选择-score(选完剩下的))。对于这道题来说，dp[i][j] = max(nums[i] - dp[i + 1][j], nums[j] - dp[i][j-1])。一共有N步，用DP的话也要把问题拆成N步对应起来，虽然对每步有很多的可能状态，但最后一步只剩下一个元素，这个元素只能是nums的每个元素，因此bottom-up。

Code

class Solution {
public:
    bool PredictTheWinner(vector<int>& nums) {
        const int n = nums.size();
        vector<vector<int>> dp(n, vector<int>(n, INT_MIN));
        for (int i = 0; i < n; ++i) {
            dp[i][i] = nums[i];
        }
        for (int l = 2; l <= n; ++l) {
            for (int i = 0; i <= n - l; ++i) {
                int j = i + l - 1;
                dp[i][j] = max(nums[i] - dp[i + 1][j], nums[j] - dp[i][j - 1]);
            }
        }

        return dp[0][n - 1] >= 0;
    }
};

时间复杂度

O(N^2)