Leetcode:486. Predict the Winner

预测赢家的动态规划算法

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本文介绍了一种通过动态规划解决两人游戏得分预测问题的方法。玩家轮流从整数数组两端取数，目标是获得比对手更高的总分。文章详细解释了动态规划公式，并给出了解决方案的具体实现。

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.

解题思路

本题为动态规划问题，DP公式为：
dp(i, j) = max(sum(i, j-1) - dp(i, j-1) + nums[j], sum(i+1, j) - dp(i+1, j) + nums[i])
因sum(i, j-1) + nums[j] = sum(i, j) = nums[i] + sum(i+1, j), 则dp(i, j) = max(sum(i, j) - dp(i, j-1), sum(i, j) - dp(i+1, j))，由此化简sum(i,j)得到dp(i, j) = max(nums[j] - dp(i, j-1), nums[i] - dp(i+1, j))

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