leetCode-1-bit and 2-bit Characters

Description:
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.

My Solution:

class Solution {
    public boolean isOneBitCharacter(int[] bits) {
        int len = bits.length;
        for(int i = 0;i < len;i++){
            if(i == len - 2){
                if(bits[i] == 1){
                    return false;
                }
                if(bits[i] == 0){
                    return true;
                }
            }
            if(bits[i] == 1){
                   String temp = bits[i] + "" + bits[i+1];
                   if(temp.equals("10") || temp.equals("11")){
                        i++;
                    }
            }
        }
        return true;
    }
}

Another Better Solution:

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

总结:我们注意到’0’和’10’,’11’是”互斥的”,即没有’01’这个干扰，所有当前位为’0’,则跳过一位，当前位为’1’，则跳过两位,最后通过i判断，如果i=bits.length-1，说明最后位为0,那么返回true,如果i=bits.length,说明最后可以返回二位的的字符，即为false