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.
Solution
有点妙的一道题。注意bits已知是以0结尾的，不用考虑以01结尾这种无法decode的情况，只需要考虑以如下几种为结尾的情况：
- 00: return true
- 010: return false
- 0110: return true
- 01110: return false
得到规律如下：统计最后一个0前面连续1的个数，如果为偶数则返回true，否则返回false。
class Solution {
public boolean isOneBitCharacter(int[] bits) {
boolean evenOnes = true;
for (int i = bits.length - 2; i >= 0 && bits[i] == 1; --i) {
evenOnes = !evenOnes;
}
return evenOnes;
}
}
