LeetCode之1-bit|LeetCode之1-bit and 2-bit Characters（Kotlin） LeetCode之1-bitand2-bitCharacters（

问题：
We have two special characters. The first character can be represented by one bit 0. The second character can be represented by two bits (10 or 11).
Now given a string represented by several bits. Return whether the last character must be a one-bit character or not. The given string will always end with a zero.
Note:
1 <= len(bits) <= 1000.
bits[i] is always 0 or 1.

方法：
遍历数组，如果当前bit为0则指针移动一位，因为0必为first character；如果当前bit为1则指针移动两位，因为1X必为second character。当遍历到数组尾端时，如果指针正好等于数组最后一个元素的下标，则返回true；否则，数组尾端必为10，返回false。
具体实现：

class IsOneBitCharacter { fun isOneBitCharacter(bits: IntArray): Boolean { var i = 0 while (i <= bits.lastIndex - 1) { if (bits[i] == 0) { i++ } else if (bits[i] == 1) { i += 2 } } if (i == bits.lastIndex) { return true } return false } }fun main(args: Array) { val array = intArrayOf(0, 0, 0, 0) val isOneBitCharacter = IsOneBitCharacter() val result = isOneBitCharacter.isOneBitCharacter(array) println("result: $result") }

有问题随时沟通
【LeetCode之1-bit|LeetCode之1-bit and 2-bit Characters（Kotlin）】具体代码实现可以参考Github