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.

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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.

题意&解题思路

给定一个非负数组，然后两个玩家轮流从数组两端取数，两个玩家都是按照最优策略去取数，最后取得数大者赢得比赛。求玩家1取到是否能赢。

令 dp[i][j] 表示玩家1在子数组[i, …, j]中能拿到的最大分数，sum[i, j] 表示数组[i, …, j]之和，那么可以得：
dp[i][j] = max(sum[i, j - 1] - dp[i][j - 1] + nums[j], sum[i + 1, j] - dp[i + 1][j] + nums[i])

又 sum[i, j - 1] + nums[j] = sum[i, j] = sum[i + 1, j] + nums[i]
故 dp[i][j] = max(sum[i, j] - dp[i][j - 1], sum[i, j] - dp[i + 1][j])

从上式可看到，右式的两项均含有sum[i, j]，我们可以考虑将其消去。

换个想法，与其保存玩家1在每个子数组中能获得的最大分数，不如直接存储玩家1和玩家2能获得的最大分数的差。比如，若玩家1获得 A 分，玩家2获得 B 分，那我们用 dp’ 来存储 A-B .

若 A = dp[i][j] ，那么 B = sum(i, j) - dp[i][j]

故 dp’[i][j] = dp[i][j] - (sum(i, j) - dp[i][j]) = 2 * dp[i][j] - sum(i, j) ，即 2 * dp[i][j] = dp’[i][j] + sum(i, j) （后面会用到）。

dp’[i][j] = dp[i][j] - ( sum(i, j) - dp[i][j] ) = 2dp[i][j] - sum(i, j)
= 2 * max( sum(i, j) - dp[i][j-1], sum(i, j) - dp[i+1][j] ) - sum(i, j)
= max(sum(i, j) - 2*dp[i][j-1], sum(i, j) - 2*dp[i+1][j] )
= max(sum(i, j) - ( dp’[i][j-1] + sum(i, j-1) ), sum(i, j) - ( dp’[i+1][j] + sum(i+1, j)))
= max(sum(i, j) - sum(i, j-1) - dp’[i][j-1], sum(i, j) - sum(i+1, j) - dp’[i+1][j])
= max(nums[j] - dp’(i, j-1), nums[i] - dp’(i+1, j))

最后可得: dp[i][j] = max(nums[j] - dp[i][j-1], nums[i] - dp[i+1][j])
此时当 dp[i][j] >= 0 时，即表示玩家1获得的分数 >= 玩家2的分数。

代码如下：

class Solution {public:    bool PredictTheWinner(vector<int>& nums) {        int n = nums.size();        int dp[n][n] = {0};        for(int i = 0; i < n; i++)dp[i][i] = nums[i];        for(int i = 1; i < n; i++){            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;    }};