486. Predict the Winner

问题描述：

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: 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.

Example 2:

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.

大致题意：player1和player2只能只能在数组的开头和末尾进行选择，最终比较两者的总分谁大谁小。

问题分析：

      根据曾经学过的矩阵的相乘运算，很容易想到用i和j记录player1再数组[i,i+1,......j]获得的最大分数。

      状态方程：dp[i][j]：表示player1在数组i,......j获得的最大总分和。

      状态转移：dp[i][j]：表示player1获得的分数情况，其上一个状态是player2选择i+1 或者j-1。

      当palyer2选的是i+1的时候，即player2在[i+1,.....j]获得了最大的分数dp[i+1][j],那么player1此时获得的分数是sum[i+1][j]-dp[i+1][j]+nums[i];

      当palyer2选的是j-1的时候，即player2在[i,.....j-1]获得了最大的分数dp[i][j-1],那么player1此时获得的分数是sum[i][j-1]-dp[i][j-1]+nums[j];

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

       终止状态:dp[0][N-1] 判断：return dp[0][N-1]>=(sum[0][N-1]-dp[0][N-1].

代码如下：

class Solution {public:    int max(int a,int b){ return a>b?a:b;    } int sum(int beginning,int ending,vector<int>& nums){ int sum=0; for(int i=beginning;i<=ending;i++){ sum+=nums[i]; } return sum; } bool PredictTheWinner(vector<int>& nums) { vector<vector<int>> dp(nums.size(),vector<int>(nums.size(),0)); for(int i=0;i<nums.size();i++){ dp[i][i]=nums[i]; } dp[0][1]=nums[0]>nums[1]?nums[0]:nums[1];  for(int r=2;r<=nums.size();++r) {     for(int i=0;i<nums.size()-r+1;++i)     {         int j=i+r-1;         dp[i][j]=max(sum(i+1,j,nums)-dp[i+1][j]+nums[i],sum(i,j-1,nums)-dp[i][j-1]+nums[j]);     } } /*for(int i=1;i<nums.size();i++){    for(int j=i+1;j<nums.size();j++){dp[i][j]=max(sum(i+1,j,nums)-dp[i+1][j]+nums[i],sum(i,j-1,nums)-dp[i][j-1]+nums[j]);  } }*/ return 2*dp[0][nums.size()-1]>=sum(0,nums.size()-1,nums); }};