zoukankan      html  css  js  c++  java
  • 717. 1-bit and 2-bit Characters--easy

    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.

    1.思考

    • 根据题意可知,遍历整个序列,若当前是0,则index++;若当前是1,则index = index + 2;
    • 直到序列最后两位,若倒数第二位为1,则index = index + 2,最后一位不是1bit,return false;
    • 若倒数第二位为0,则看最后一位是否为0,是则return true,否则return false;
    • 即当解码跳出循环时为indexlen-1,且bits[index]0,return true;否则,return false。

    2.实现
    Runtime: 4ms(100%)
    Memory: 8.5MB(100%)

    class Solution {
    public:
        bool isOneBitCharacter(vector<int>& bits) {
            int len = bits.size();
            int i = 0;
            while(i<len-1){
                if(bits[i]==1)
                    i++; 
                i++;
            }
            
            if(i==len-1 && bits[i]==0)
                return true;
            else
                return false;
        }
    };
    
  • 相关阅读:
    dd的用法
    od的用法
    Windows 7安装Oracle 10g的方法
    Ubuntu下的iptux和Windows下的飞秋互传文件
    c++ 12
    c++ 11
    c++ 10
    c++ 09
    c++ 08
    c++ 07
  • 原文地址:https://www.cnblogs.com/xuyy-isee/p/10654398.html
Copyright © 2011-2022 走看看