leetcode 717. 1-bit and 2-bit Characters

最新推荐文章于 2025-05-18 07:15:00 发布

转载最新推荐文章于 2025-05-18 07:15:00 发布 · 47 阅读

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原文链接：http://www.cnblogs.com/pk28/p/7761081.html

本文介绍了一种算法，用于判断给定比特串的最后一个字符是否为一比特长的特殊字符。通过遍历比特串并根据当前位的值跳过相应长度的字符，可以确定最后一个字符的长度。

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.

直接判断就好了。对于当前位，是1的话直接条到i+2是0的话不用管，然后看最后一位是不是0

class Solution {
public:
    bool isOneBitCharacter(vector<int>& bits) {
        for (int i = 0; i < bits.size(); ++i) {
            if (bits[i] == 1) {
                if (i + 1 < bits.size()) {
                    if (bits[i+1] == 0 || bits[i+1] == 1) {
                        i++;continue;
                    }
                    else {
                        return false;
                    }
                } else {
                    return false;
                }
            }
            else {
                if (i == bits.size()-1) return true;
            }
        }
        return false;
    }
};

转载于:https://www.cnblogs.com/pk28/p/7761081.html