486. Predict the Winner

题目：

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.

解题思路：

两个选手轮流从任意一边取一个score,直到没有score为止。计算拿到的score总值，选手1的数值高于或等于选手2的数值，返回TRUE，否则返回FALSE。
这个题的关键公式为dp[i] = max(nums[i] + sum(i+1,i+k) - dp[i+1] , nums[i+k] + sum(i,i+k-1) - dp[i]) i 从后向前遍历。这里有一个问题，dp[i+1]这个值在计算dp[i+1]的时候被改变了，但是在计算dp[i]之中又用到了，所以在改变之前需要存储这个值。
具体代码如下：

class Solution {  public:      bool PredictTheWinner(vector<int>& nums) {          int len = nums.size();          int * sum = new int[len + 1];          int * dp = new int[len];          sum[0] = 0;          for(int i = 0; i <= len; i++) {              sum[i] = sum[i-1] + nums[i-1];          }          for(int i = 0; i < len; i++) {              dp[i] = nums[i];          }          int temp = 0;          for(int j = 1; j < len; j++) {              temp = dp[len-j];              for(int i = len - j - 1; i >= 0; i--) {                  int temp1 = dp[i];                  dp[i] = max(nums[i] + (sum[i+j+1] - sum[i+1]) - temp,nums[i+j] + (sum[i+j] - sum[i]) - dp[i]);                  temp = temp1;              }          }          if(dp[0]*2 >= sum[len])              return true;          else              return false;      }  };