1-bit_and_2-bit_Characters.md

717. 1-bit and 2-bit Characters

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.

Example 1:

Input: bits = [1, 0, 0]
Output: True
Explanation: 
The only way to decode it is two-bit character and one-bit character. So the last character is one-bit character.

Example 2:

Input: bits = [1, 1, 1, 0]
Output: False
Explanation: 
The only way to decode it is two-bit character and two-bit character. So the last character is NOT one-bit character.

Note:

• 1 <= len(bits) <= 1000.
• bits[i] is always 0 or 1.

Code:

class Solution {
    public boolean isOneBitCharacter(int[] bits) {
        int len = bits.length;
        boolean res = true;
        int pos = 0;
        while (pos < len - 1) {
            if (pos == len - 2 && bits[pos] == 1) res = false; //找到最后一位是两位字符的情况
            if (bits[pos] == 1) {
                //如果当前元素为一，指针移两位
                pos += 2;
            } else if (bits[pos] == 0) {
                //如果当前元素为零，指针移一位
                pos += 1;
            }
        }
        return res;
    }
}

---

• 该题的解题思路是找到数组中最后一位是两位字符的情况，如果倒数第二位能和最后一位组合成1，0则返回false。
• 遍历数组，如果遇到1，则指针加二；如果遇到零，则指针加一。

Enhanced Code:

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

---

• 改进算法的优势在于：在遍历时终止条件是判断倒数第三位元素，如果倒数第三位元素是1，则和倒数第二位可组合为两位字符的情况，指针接下来指向最后一位；如果倒数第三位元素是0，则是单字符的解码，指针接下来不会指向最后一位。
• 在最后遍历结束时，判断指针是否指向最后一位。