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.

就是一个自底向上的动态规划。并且始终只有一个子问题，所以复杂度 。

```
class Solution {
private int[] local = new int[0];
private int len = 0;
public boolean isOneBitCharacter(int[] bits) {
local = bits;
len = bits.length;
return dp(0);
}
private boolean dp(int index) {
if (index == len) { return false; }
if (index == len - 1) { return true; } // assertion: always end with 0
if (local[index] == 0) {
return dp(index + 1);
} else {
return dp(index + 2);
}
}
}
```

```
class Solution {
public boolean isOneBitCharacter(int[] bits) {
int cur = 0;
while (cur < bits.length - 1) {
cur = cur + ((bits[cur] == 0)? 1 : 2);
}
return cur == bits.length - 1;
}
}
```