Alex and Lee play a game with piles of stones. There are an even number of piles arranged in a row, and each pile has a positive integer number of stones piles[i].
The objective of the game is to end with the most stones. The total number of stones is odd, so there are no ties.
Alex and Lee take turns, with Alex starting first. Each turn, a player takes the entire pile of stones from either the beginning or the end of the row. This continues until there are no more piles left, at which point the person with the most stones wins.
Assuming Alex and Lee play optimally, return True if and only if Alex wins the game.
Example 1:
Input: [5,3,4,5] Output: true Explanation: Alex starts first, and can only take the first 5 or the last 5. Say he takes the first 5, so that the row becomes [3, 4, 5]. If Lee takes 3, then the board is [4, 5], and Alex takes 5 to win with 10 points. If Lee takes the last 5, then the board is [3, 4], and Alex takes 4 to win with 9 points. This demonstrated that taking the first 5 was a winning move for Alex, so we return true.
思路：先求出子問題即每兩個石堆之間的差值，即Alex所能得到的比Lee多的部分，然後求連續三個石堆，四個。。，依次遞推，最後求出原問題的解，dp[i][j]表示兩石堆間Alex比Lee所能獲得的多出來的部分。最後判斷這個總差值是否大於0，若大於0則判定Alex贏。（注意dp[i][i]的初始化）
class Solution {
public:
bool stoneGame(vector<int>& piles) {
int n = piles.size();
vector<vector<int> > dp(n,vector<int>(n,0));
for(int i=0;i<n;i++){
dp[i][i] = piles[i];
}
for(int i=1;i<n;i++){ //i is the distance of the neighbor
for(int j=0;j<n-i;j++){
dp[j][j+i] = max(abs(dp[j+1][j+i]-piles[j]),abs(dp[j][j+i-1]-piles[j]));
}
}
if(dp[0][n-1]>0) return true;
else return false;
}
};