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： 遍歷數組，若是當前bit爲0則指針移動一位，由於0必爲first character；若是當前bit爲1則指針移動兩位，由於1X必爲second character。當遍歷到數組尾端時，若是指針正好等於數組最後一個元素的下標，則返回true；不然，數組尾端必爲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")
}
複製代碼

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具體代碼實現能夠參考Github