1-bit and 2-bit Characters(leetcode)

1-bit and 2-bit Characters

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题目

leetcode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: 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.

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解决

遍历给定的字符串：

• 若遇到bits[i] == 0，跳过，不做任何处理
• 若遇到bits[i] == 1，判断是否处在最后一位(无法构建题目中的character)或者倒数第二位(自动与最后一位的bits[n - 1]构成character)。若满足i == n - 1 || i + 1 == n - 1，都不能使得最后的bit变成一位的character。若不满足，接着判断bits[i + 2]，直到循环结束。

class Solution {
public:
    bool isOneBitCharacter(vector<int>& bits) {
        int num = bits.size();
        for (int i = 0; i < num; i++) {
            if (bits[i] == 1) {
                if (i == num - 1 || i + 1 == num - 1) {
                    return false;
                }
                i++;
            }
        }
        return true;
    }
};