717 1-bit and 2-bit Characters

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.
> ```

题目大意：给定一个位表示的数组，问是否最后一位可以one-bit表示，其中one-bit是1，two bits是10或者11.

思路：题目简单来说，就是给定三种数字，分别是11,10,0问是否可以通过这些数组成给定的数组，并且数组的最后一位用0（是ont bit的0不是两位的10）表示。因为两位的表示都是以1开头的，而一位的表示没有1，所以设置一个记录位置i，初始i为0，当idx<数组长度-1时，进行遍历，idx遇到1（因为此时代表2位）时加2，遇到0（此时代表一位）时候加1，看最后的idx是否等于数组长度-1。

package Array;

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class Solution {
public boolean isOneBitCharacter(int[] bits) {
int n = bits.length;
int idx = 0;
while (idx < n - 1) {
//            遇到两位的情况
if (bits[idx] == 1) {
idx += 2;
} else {
//                一位的情况
idx++;
}

}
return idx == n - 1;
}
}