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:

bits[i] is always 0 or 1.

1.從頭掃描整個整陣列，0可以單獨一個字元表示，1必須兩個字元表示。

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

2.從檢查陣列末尾

如果是00則肯定true，因為沒有兩個字元可以表示00；

如果是10則統計連續的1的個數，連續1的個數為奇數則為false 個數為偶數則為true。

class Solution {
    public boolean isOneBitCharacter(int[] bits) {
        int ones = 0;
        //Starting from one but last, as last one is always 0.
        for (int i = bits.length - 2; i >= 0 && bits[i] != 0 ; i--) { 
            ones++;
        }
        if (ones % 2 > 0) return false; 
        return true;
    }
}