交替取值判断第一个人能否赢过第二个人 Predict the Winner-优快云博客

本文介绍了一种通过算法预测两位玩家在游戏中得分胜负的方法。玩家轮流从数组两端取数，最终得分高者获胜。文章提供了三种解决方案，包括递归方法、递归进化版以及高效的动态规划方法。

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问题：

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.

解决：

【题意】给定一个正整数数组，选手1从数组的头部或者尾部选择一个数，选手2从剩下部分的头部或尾部选择一个数，循环往复，直到该数组中的数都被取完。判断选手1取的数的和值是否大于选手2.

① 两人依次拿，如果Player1赢，则Player1拿的>Player2拿的。我们把Player1拿的视为"+"，把Player2拿的视为"-"，如果最后结果大于等于0则Player1赢。

递归方法解决：

class Solution { //67ms
public boolean PredictTheWinner(int[] nums) {
return dfs(nums,0,nums.length - 1) >= 0;
}
public int dfs(int[] nums,int start,int end){
if (start == end) {
return nums[start];
}else {
return Math.max(nums[start] - dfs(nums,start + 1,end),nums[end] - dfs(nums,start,end - 1));
}
}
}

② 递归进化版，记录中间状态。

class Solution {//91ms
int[][] dp;
public boolean PredictTheWinner(int[] nums) {
dp = new int[nums.length][nums.length];
return dfs(nums,0,nums.length - 1) >= 0;
}
public int dfs(int[] nums,int start,int end){
if (dp[start][end] == 0){
if (start == end) return nums[start];
else{
return Math.max(nums[start] - dfs(nums,start + 1,end),nums[end] - dfs(nums,start,end - 1));
}
}
return dp[start][end];
}
}

③ 动态规划。dp[i][j] = Math.max(nums[i] - dp[i + 1][j], nums[j] - dp[i][j - 1]);

class Solution { //7ms
public boolean PredictTheWinner(int[] nums) {
if (nums == null || nums.length == 0) return false;
int len = nums.length;
int[][] dp = new int[len][len];
for (int i = 0;i < len;i ++){
dp[i][i] = nums[i];
}
for (int i = len - 2;i >= 0;i --){
for (int j = i + 1;j < len;j ++){
dp[i][j] = Math.max(nums[i] - dp[i + 1][j],nums[j] - dp[i][j - 1]);
}
}
return dp[0][len - 1] >= 0;
}
}