LN : leetcode 486 Predict the Winner


lc 486 Predict the Winner


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.

DP Accepted

dp[i][j]代表在i與j之間,先發者最多能比后發者多拿的分數。動態轉移方程dp[j][j+i] = max(nums[j+i]-dp[j][j+i-1], nums[j]-dp[j+1][j+i])代表最大值要么是取頭並加上去頭至尾部的dp值,要么是取尾並加上頭至去尾的dp值,注意i、j的取值范圍。

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

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