算法分析与设计课程12——486. Predict the Winner

一、题目描述【原题链接】

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.

Input: [1, 5, 2]Output: FalseExplanation: 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.

Input: [1, 5, 233, 7]Output: TrueExplanation: 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.

二、分析

使用动态规划解决该问题

另dp[i][j]为：从 i~j 范围【先手玩家】能够取到的总数值 减去 【后手玩家】能取到的总数值。

当dp[0][nums.length - 1] > 0时代表Player1赢，否则输。

对于每个dp[i][j]，取下面两种情况最大值
①从左边取，则有：dp[i][j]=nums[i] - dp[i+1][j]（假设玩家1从左边先手取后，玩家2在i+1~j范围先手取，下同）
②从右边取，则有：dp[i][j]=nums[j] - dp[i][j - 1]

三、源代码

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