[Leetcode] 717. 1-bit and 2-bit Characters 解题报告

最新推荐文章于 2019-09-06 19:42:05 发布

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本文介绍了一种特殊字符编码方案，并提供了一个算法来判断给定字符串的最后一个字符是否为一比特字符。通过对输入比特序列的解析，可以确定每个字符是由一个比特还是两个比特组成。

题目：

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.
思路：

通过观察可知，这种特殊的字符只有三种形式：10,11和0，所以当我们遇到以1开头的bit时，它一定属于两个bit构成的字符；遇到以0开头的bit的时候，它一定属于1个bit构成的字符。这样让i从0开始一直循环到bits的最后一个字符，如果i就是最后一个字符的索引，说明最后一个字符就是1bit的，否则说明是2bits的。

代码：
```
class Solution {
public:
    bool isOneBitCharacter(vector<int>& bits) {
        int len = bits.size();
        int i = 0;
        while (i < len - 1) {
            if (bits[i] == 0) {
                ++i;
            }
            else {
                i += 2;
            }
        }
        return i == len - 1;
    }
};
```