LeetCode.717 1-bit and 2-bit Characters

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: TrueExplanation: 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: FalseExplanation: 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.

分析：

class Solution {    public boolean isOneBitCharacter(int[] bits) {        //给定数组，最后一位是0，其中规定1bit只能是0，2bit是10或者11。返回该数组能只有这两种bit组成        //思路：分两种情况，2bit为10和11，1bit为0。那么只需要统计2bit出现的次数和1bit出现的次数        //最后将次数乘以对应的长度和总的数组长度对比即可                int even=0,odd=0;        for(int i=0;i<bits.length;i++){            if(i<bits.length-1&&bits[i]==1&&(bits[i+1]==0||bits[i+1]==1)){                //为2bit的情况                i++;                even++;            }else if(i<bits.length-1&&bits[i]==0){                odd++;            }                        if(i==bits.length-1&&bits[i]==0){                odd++;            }        }                return even*2+odd==bits.length?true:false;            }}