LeetCode之1 bit and 2 bit Characters（Kotlin）

問題： 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.

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方法： 遍歷陣列，如果當前bit為0則指標移動一位，因為0必為first character；如果當前bit為1則指標移動兩位，因為1X必為second character。當遍歷到陣列尾端時，如果指標正好等於陣列最後一個元素的下標，則返回true；否則，陣列尾端必為10，返回false。

具體實現：

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<String>) {
    val array = intArrayOf(0, 0, 0, 0)
    val isOneBitCharacter = IsOneBitCharacter()
    val result = isOneBitCharacter.isOneBitCharacter(array)
    println("result: $result")
}
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具體程式碼實現可以參考Github