Alice and Bob take turns playing a game, with Alice starting first.
\r\n\r\nThere are n stones arranged in a row. On each player's turn, while the number of stones is more than one, they will do the following:
x > 1, and remove the leftmost x stones from the row.The game stops when only one stone is left in the row.
\r\n\r\nThe score difference between Alice and Bob is (Alice's score - Bob's score). Alice's goal is to maximize the score difference, and Bob's goal is the minimize the score difference.
Given an integer array stones of length n where stones[i] represents the value of the ith stone from the left, return the score difference between Alice and Bob if they both play optimally.
\r\n
Example 1:
\r\n\r\n\r\nInput: stones = [-1,2,-3,4,-5]\r\nOutput: 5\r\nExplanation:\r\n- Alice removes the first 4 stones, adds (-1) + 2 + (-3) + 4 = 2 to her score, and places a stone of\r\n value 2 on the left. stones = [2,-5].\r\n- Bob removes the first 2 stones, adds 2 + (-5) = -3 to his score, and places a stone of value -3 on\r\n the left. stones = [-3].\r\nThe difference between their scores is 2 - (-3) = 5.\r\n\r\n\r\n
Example 2:
\r\n\r\n\r\nInput: stones = [7,-6,5,10,5,-2,-6]\r\nOutput: 13\r\nExplanation:\r\n- Alice removes all stones, adds 7 + (-6) + 5 + 10 + 5 + (-2) + (-6) = 13 to her score, and places a\r\n stone of value 13 on the left. stones = [13].\r\nThe difference between their scores is 13 - 0 = 13.\r\n\r\n\r\n
Example 3:
\r\n\r\n\r\nInput: stones = [-10,-12]\r\nOutput: -22\r\nExplanation:\r\n- Alice can only make one move, which is to remove both stones. She adds (-10) + (-12) = -22 to her\r\n score and places a stone of value -22 on the left. stones = [-22].\r\nThe difference between their scores is (-22) - 0 = -22.\r\n\r\n\r\n
\r\n
Constraints:
\r\n\r\nn == stones.length2 <= n <= 105-104 <= stones[i] <= 104Alice 和 Bob 玩一个游戏，两人轮流操作， Alice 先手 。
\n\n总共有 n 个石子排成一行。轮到某个玩家的回合时，如果石子的数目 大于 1 ，他将执行以下操作：
x > 1 ，并且 移除 最左边的 x 个石子。当只剩下 一个 石子时，游戏结束。
\n\nAlice 和 Bob 的 分数之差 为 (Alice 的分数 - Bob 的分数) 。 Alice 的目标是 最大化 分数差，Bob 的目标是 最小化 分数差。
给你一个长度为 n 的整数数组 stones ，其中 stones[i] 是 从左边起 第 i 个石子的价值。请你返回在双方都采用 最优 策略的情况下，Alice 和 Bob 的 分数之差 。
\n\n
示例 1：
\n\n输入：stones = [-1,2,-3,4,-5]\n输出：5\n解释：\n- Alice 移除最左边的 4 个石子，得分增加 (-1) + 2 + (-3) + 4 = 2 ，并且将一个价值为 2 的石子放在最左边。stones = [2,-5] 。\n- Bob 移除最左边的 2 个石子，得分增加 2 + (-5) = -3 ，并且将一个价值为 -3 的石子放在最左边。stones = [-3] 。\n两者分数之差为 2 - (-3) = 5 。\n\n\n
示例 2：
\n\n输入：stones = [7,-6,5,10,5,-2,-6]\n输出：13\n解释：\n- Alice 移除所有石子，得分增加 7 + (-6) + 5 + 10 + 5 + (-2) + (-6) = 13 ，并且将一个价值为 13 的石子放在最左边。stones = [13] 。\n两者分数之差为 13 - 0 = 13 。\n\n\n
示例 3：
\n\n输入：stones = [-10,-12]\n输出：-22\n解释：\n- Alice 只有一种操作，就是移除所有石子。得分增加 (-10) + (-12) = -22 ，并且将一个价值为 -22 的石子放在最左边。stones = [-22] 。\n两者分数之差为 (-22) - 0 = -22 。\n\n\n
\n\n
提示：
\n\nn == stones.length2 <= n <= 105-104 <= stones[i] <= 104\\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\"],\"java\":[\"Java\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u88ab\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u5305\\u542b Pair \\u7c7b: https:\\/\\/docs.oracle.com\\/javase\\/8\\/javafx\\/api\\/javafx\\/util\\/Pair.html <\\/p>\"],\"python\":[\"Python\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982\\uff1aarray<\\/a>, bisect<\\/a>, collections<\\/a>\\u3002\\u5982\\u679c\\u60a8\\u9700\\u8981\\u4f7f\\u7528\\u5176\\u4ed6\\u5e93\\u51fd\\u6570\\uff0c\\u8bf7\\u81ea\\u884c\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u6ce8\\u610f Python 2.7 \\u5c06\\u57282020\\u5e74\\u540e\\u4e0d\\u518d\\u7ef4\\u62a4<\\/a>\\u3002 \\u5982\\u60f3\\u4f7f\\u7528\\u6700\\u65b0\\u7248\\u7684Python\\uff0c\\u8bf7\\u9009\\u62e9Python 3\\u3002<\\/p>\"],\"c\":[\"C\",\" \\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u5982\\u60f3\\u4f7f\\u7528\\u54c8\\u5e0c\\u8868\\u8fd0\\u7b97, \\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 uthash<\\/a>\\u3002 \\\"uthash.h\\\"\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5bfc\\u5165\\u3002\\u8bf7\\u770b\\u5982\\u4e0b\\u793a\\u4f8b:<\\/p>\\r\\n\\r\\n 1. 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