leetcode-717-1-bit-and-2-bit-Characters

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描述

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:

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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:

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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.

分析

输入是一个数组，其中的元素是 10，11， 0，判断最后一个元素是否为 0。注意到 0 长度是1，10，11 的长度都是2，一次遍历即可。时间复杂度为 O(n), 空间复杂度为 O(n)

解决方案1(Java)

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

相关问题

(M) Gray Code

题目来源