717. 1-bit and 2-bit Characters-优快云博客

本文介绍了一种特殊字符编码方式，并提供了三种不同的算法来判断给定字符串是否以一比特字符结束。通过遍历检查和递归复制等方法，确保了算法的正确性和效率。

题目描述：

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.

思路一：

class Solution {
    public boolean isOneBitCharacter(int[] bits) {
        int i = 0;
        while(i < bits.length - 1)
        {
            if (bits[i] == 0) i++;
            else i += 2;
        }
        return  i == bits.length - 1;
    }
}

思路二：

只检查数列的最后一部分。

*如果最后两个数都为0，则true

*如果倒数第二个数为1，则取决于1的个数。1的个数为偶数，则true；1的个数为奇数，则false。

class Solution {
    public boolean isOneBitCharacter(int[] bits) {
        int i = bits.length - 2;
        int ones = 0;
        while (i >= 0 && bits[i] != 0)
        {
            ones++;
            i--;
        }
        return ones % 2 == 0;
    }
}

思路三：

class Solution {
    public boolean isOneBitCharacter(int[] bits) {
        return bits.length <= 2 ? bits[0] == 0 : isOneBitCharacter(Arrays.copyOfRange(bits, bits[0] + 1, bits.length));
    }
}