A frog is crossing a river. The river is divided into some number of units, and at each unit, there may or may not exist a stone. The frog can jump on a stone, but it must not jump into the water.
\n\nGiven a list of stones positions (in units) in sorted ascending order, determine if the frog can cross the river by landing on the last stone. Initially, the frog is on the first stone and assumes the first jump must be 1 unit.
If the frog's last jump was k units, its next jump must be either k - 1, k, or k + 1 units. The frog can only jump in the forward direction.
\n
Example 1:
\n\n\nInput: stones = [0,1,3,5,6,8,12,17]\nOutput: true\nExplanation: The frog can jump to the last stone by jumping 1 unit to the 2nd stone, then 2 units to the 3rd stone, then 2 units to the 4th stone, then 3 units to the 6th stone, 4 units to the 7th stone, and 5 units to the 8th stone.\n\n\n
Example 2:
\n\n\nInput: stones = [0,1,2,3,4,8,9,11]\nOutput: false\nExplanation: There is no way to jump to the last stone as the gap between the 5th and 6th stone is too large.\n\n\n
\n
Constraints:
\n\n2 <= stones.length <= 20000 <= stones[i] <= 231 - 1stones[0] == 0stones is sorted in a strictly increasing order.一只青蛙想要过河。 假定河流被等分为若干个单元格，并且在每一个单元格内都有可能放有一块石子（也有可能没有）。 青蛙可以跳上石子，但是不可以跳入水中。
\n\n给你石子的位置列表 stones（用单元格序号 升序 表示）， 请判定青蛙能否成功过河（即能否在最后一步跳至最后一块石子上）。开始时， 青蛙默认已站在第一块石子上，并可以假定它第一步只能跳跃 1 个单位（即只能从单元格 1 跳至单元格 2 ）。
如果青蛙上一步跳跃了 k 个单位，那么它接下来的跳跃距离只能选择为 k - 1、k 或 k + 1 个单位。 另请注意，青蛙只能向前方（终点的方向）跳跃。
\n\n
示例 1：
\n\n\n输入：stones = [0,1,3,5,6,8,12,17]\n输出：true\n解释：青蛙可以成功过河，按照如下方案跳跃：跳 1 个单位到第 2 块石子, 然后跳 2 个单位到第 3 块石子, 接着 跳 2 个单位到第 4 块石子, 然后跳 3 个单位到第 6 块石子, 跳 4 个单位到第 7 块石子, 最后，跳 5 个单位到第 8 个石子（即最后一块石子）。\n\n
示例 2：
\n\n\n输入：stones = [0,1,2,3,4,8,9,11]\n输出：false\n解释：这是因为第 5 和第 6 个石子之间的间距太大，没有可选的方案供青蛙跳跃过去。\n\n
\n\n
提示：
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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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