LeetCode 486. Predict the Winner

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.

分析

dp[i][j]表示nums数组中i～j下标间player1能够获得的分数-player2能够获得的分数～最后dp[0][len-1]的正负性即可表明player1是否能赢～,关键是自底而上从长度由小到大的来更新dp以及转移方程dp[i][j]=max(nums[i]-dp[i+1][j],nums[j]-dp[i][j-1]).其中nums[i]-dp[i+1][j]表示player1先选了第一个数，而nums[j]-dp[i][j-1]表示player1先选了最后一个数。

class Solution {
public:
    bool PredictTheWinner(vector<int>& nums) {
        int n = nums.size();
        vector<vector<int>> dp(n, vector<int>(n, 0));
        for (int i = 0; i < n; i++) 
            dp[i][i] = nums[i];
        for (int k = 1; k < n; k++) {
            for (int i = 0; i+k < n; i++) {
                int j = i+k;
                int t1 = nums[i] - dp[i+1][j];
                int t2 = nums[j] - dp[i][j-1];
                dp[i][j] = max(t1, t2);
            }
        }
        return dp[0][n-1] >= 0;
    }
};