717. 1-bit and 2-bit Characters

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于 2018-04-13 17:23:01 发布

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分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/rockywallet/article/details/79932041

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本文介绍了一种用于识别特殊字符串是否以一比特字符结尾的算法。该算法通过判断字符串中特定模式来确定最后一个字符是否为单比特。核心在于检查字符串尾部的0前是否有偶数个连续的1。

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原题连接

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.

思路：若最后一位是1，则结果必为false；
若最后一位是0，只需要判断前边有多少个连续的1，若为偶数个，则最后一个0不用和前边的1凑对，否则必须凑对。

AC代码：

class Solution {
public:
    bool isOneBitCharacter(vector<int>& bits) {
        int length = bits.size();
        bool ok = true;
        if(bits[length-1]==0)
        {
            int pos = length-2;
            int cnt = 0;
            while(pos>=0&&bits[pos--]==1) cnt++;
            if(cnt&1) ok = false;
        }
        else 
            ok = false;
        return ok;
    }
};