Comment on page

# 717.1-bit-and-2-bit-Characters

## 题目描述

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.

## 代码

### Approach #1 Increment Pointer

When reading from the `i`-th position, if `bits[i] == 0`, the next character must have 1 bit; else if `bits[i] == 1`, the next character must have 2 bits. We increment our read-pointer `i` to the start of the next character appropriately. At the end, if our pointer is at `bits.length - 1`, then the last character must have a size of 1 bit.
class Solution {
public boolean isOneBitCharacter(int[] bits) {
int i = 0;
while (i < bits.length - 1) {
i += bits[i] + 1;
}
return i == bits.length - 1;
}
}

### Approach #2 Greedy

The second-last `0` must be the end of a character (or, the beginning of the array if it doesn't exist). Looking from that position forward, the array `bits` takes the form `[1, 1, ..., 1, 0]` where there are zero or more `1`'s present in total. It is easy to show that the answer is `true` if and only if there are an even number of ones present.
In our algorithm, we will find the second last zero by performing a linear scan from the right. We present two slightly different approaches below.
class Solution {
public boolean isOneBitCharacter(int[] bits) {
int i = bits.length - 2;
while (i >= 0 && bits[i] > 0) i--;
return (bits.length - i) % 2 == 0;
}
}