LeetCode 717. 1-bit and 2-bit Characters

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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.

在这道题中，给定了两类特殊字符，一类是一位字符：0；一类是两位字符：10或11.

现给了一个由0和1组成的、且最后一位是0的数组，判断能否将其正确分割，且最后一位是单个字符0.

思路：如果出现1，则必须跟后面的0或1组成一个两位字符。因此从前往后遍历，如果bits[i] = 1，则将bits[i + 1]改为1；循环结束后，如果数组最后一位是1，则一定是一个两位字符，如果是0，则是一个一位字符。具体代码如下：

public class Solution {
    public boolean isOneBitCharacter(int[] bits){
        for(int i = 0; i < bits.length - 1; i++){
            if(bits[i] == 1){
                bits[i + 1] = 1;
                i++;
            }
        }
        return bits[bits.length - 1] == 0; 
    }

LeetCode 717. 1-bit and 2-bit Characters

标签：==   public   turn   out   cte   nbsp   not   return   因此   

原文地址：http://www.cnblogs.com/xyyuan/p/7800795.html