LeetCode-717. 1-bit and 2-bit Characters-优快云博客

本文链接：https://blog.youkuaiyun.com/ZHOUBEISI/article/details/79431413

本文介绍了一种判断字符串是否以一比特字符结尾的算法。通过两种方法实现：一是从前向后遍历，二是从后向前统计。适用于由一比特字符'0'和两比特字符'10'/'11'组成的字符串。

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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.

1.从头扫描整个整数组，0可以单独一个字符表示，1必须两个字符表示。

class Solution {
    public boolean isOneBitCharacter(int[] bits) {
        int lens=bits.length,i;
        for(i=0;i<lens-1;) {
        	if(bits[i]==0)i++;
        	if(bits[i]==1)i+=2;
        }
    	return i==lens-1;
    }
}

2.从检查数组末尾

如果是00则肯定true，因为没有两个字符可以表示00；

如果是10则统计连续的1的个数，连续1的个数为奇数则为false 个数为偶数则为true。

class Solution {
    public boolean isOneBitCharacter(int[] bits) {
        int ones = 0;
        //Starting from one but last, as last one is always 0.
        for (int i = bits.length - 2; i >= 0 && bits[i] != 0 ; i--) { 
            ones++;
        }
        if (ones % 2 > 0) return false; 
        return true;
    }
}