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 <= 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.
Solution
DP, O(n ^ 2), S(n ^ 2)
The idea behind the recursive approach is simple. The two players Player 1 and Player 2 will be taking turns alternately. For the Player 1 to be the winner, we need scorePlayer_1≥scorePlayer_2. Or in other terms, scorePlayer_1?scorePlayer_2≥0.
Thus, for the turn of Player 1, we can add its score obtained to the total score and for Player 2's turn, we can substract its score from the total score. At the end, we can check if the total score is greater than or equal to zero(equal score of both players), to predict that Player 1 will be the winner.
We can omit the use of turnturn to keep a track of the player for the current turn. To do so, we can make use of a simple observation. If the current turn belongs to, say Player 1, we pick up an element, say xx, from either end, and give the turn to Player 2. Thus, if we obtain the score for the remaining elements(leaving xx), this score, now belongs to Player 2. Thus, since Player 2 is competing against Player 1, this score should be subtracted from Player 1's current(local) score(xx) to obtain the effective score of Player 1 at the current instant.
class Solution {
public boolean PredictTheWinner(int[] nums) {
int n = nums.length;
int[][] dp = new int[n][n]; // dp[i][j] = current player score - other player score starting from i and end with j
for (int s = n - 1; s >= 0; --s) {
for (int e = s; e < n; ++e) {
int a = nums[s] - (s < e ? dp[s + 1][e] : 0);
int b = nums[e] - (s < e ? dp[s][e - 1] : 0);
dp[s][e] = Math.max(a, b);
}
}
return dp[0][n - 1] >= 0;
}
}
變種題：拿卡片游戲
每張卡片都有一個值，給定一堆卡片從一頭拿茂洒，每次可以拿一到三張照皆，兩人輪流拿豆励，求最高得分∶考慮卡片值為負的情況辣之。
