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.
Tags: Array
思路 1
观察可以发现，其实每个编码数组都只有一种解码方式，可以直接从头开始解码。
Java:
public boolean isOneBitCharacter(int[] bits) {
int i = 0;
while (i < bits.length - 1) {
i += bits[i] + 1;
}
return i == bits.length - 1;
}
思路 2
再观察我们可以发现，0肯定是一个字符编码的结尾。且题目指出所给的编码数组的最后一个编码肯定为0，那么我们只要找到倒数第二个0的位置即可，因为这两个0之间的1的个数为偶数个时，当且仅当最后一个字符的编码为一个单独的0。
Java:
public boolean isOneBitCharacter(int[] bits) {
int i = bits.length - 2;
while (i >= 0 && bits[i] != 0) i--;
return ((bits.length - i) & 1) == 0;
}
结语
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