717. 1-bit and 2-bit Characters最后一位数是否为0

［抄题］：

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.

[暴力解法]：

时间分析：

空间分析：

[优化后]：

时间分析：

空间分析：

［奇葩输出条件］：

［奇葩corner case］：

［思维问题］：

没见过，不会

［一句话思路］：

正常操作后判断位数能不能对上

［输入量］：空：正常情况：特大：特小：程序里处理到的特殊情况：异常情况（不合法不合理的输入）：

［画图］：

［一刷］：

用while循环，因为最后要差也就之差了一位数

［二刷］：

［三刷］：

［四刷］：

［五刷］：

[五分钟肉眼debug的结果]：

［总结］：

正常操作后判断位数能不能对上

［复杂度］：Time complexity: O(n) Space complexity: O(1)

［英文数据结构或算法，为什么不用别的数据结构或算法］：

［关键模板化代码］：

直接用能不能对上来返回：

//return, check
        return i == n - 1;

［其他解法］：

［Follow Up］：

［LC给出的题目变变变］：

[代码风格] ：

class Solution {
    public boolean isOneBitCharacter(int[] bits) {
        //ini
        int n = bits.length;
        int i = 0;
        
        //cc
        if (bits == null || n == 0) {
            return false;
        }
        
        //while loop
        while (i < n - 1) {
            if (bits[i] == 0) {
                i++;
            }else {
                i += 2;
            }
        }
        
        //return, check
        return i == n - 1;
    }
}

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