Alex and Lee play a game with piles of stones. There are an even number ofpilesarranged in a row, and each pile has a positive integer number of stonespiles[i].

The objective of the game is to end with the moststones. The total number of stones is odd, so there are no ties.

Alex and Lee take turns, with Alex starting first. Each turn, a playertakes 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, returnTrueif 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.
 

Note:

1. 2 <= piles.length <= 500
2. piles.lengthis even.
3. 1 <= piles[i] <= 500
4. sum(piles)is odd.

class Solution {
    public boolean stoneGame(int[] piles) {
        int alex = 0, lee = 0;
        int left = 0, right = piles.length - 1;
        while(left < right) {
            int alexcur = Math.max(piles[left], piles[right]);
            alex 
 
+= alexcur;
            if(piles[left] == alexcur) {
                lee += piles[right];
            }
            else lee += piles[left];
            left++;
            right--;
        }
        System.out.println(alex);
        return alex > lee;
    }
}

1. 啊？？？後來覺得這麼想錯了，是每個人從首尾選一個，alex先選，但為什麼還ac了？？？

return true;

2. 牛逼之直接return true，為啥？

我們知道一共有odd number總數的石頭，分為even堆，那我們要是分一半，肯定會出現sum(odd index) 》 sum（even index）或者相反，所以我們每次讓alex選odd or even index的不就行了？