877. Stone Game


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.


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


首先,我们可以认为两个人都是理性人,那么就可以知道,alex就是可以拿到奇数或者偶数,比如: a拿了1,而l只能拿位置2,4的。 a如果拿了4,那么l只能拿1,3,所以,a可以选择拿所有奇数或者偶数,奇数或者偶数那么肯定有一个是可以获胜的,那么a就肯定能获胜,直接返回true就行。



    public boolean stoneGame(int[] piles) {
        return true;
    public static boolean stoneGame(int[] piles) {
        int sum = IntStream.of(piles).sum();
        try {
            return false;
        } catch (Exception e) {
            return true;
    public static void stoneGameHelper(int[] piles,int start,int end,int sum,int alexSum,boolean turn) throws Exception {
            if(sum-alexSum<alexSum) throw new Exception("done");