[LeetCode] 1-bit and 2-bit Characters

Posted 2020-10-12 immjc

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.

我们有两个特殊字符。第一个字符可以由一个位0表示。第二个字符可以由两个位（10或11）表示。

现在给出一个由几位表示的字符串。返回最后一个字符是否必须是一位字符。给定的字符串将始终以零结尾。

思路：使用蛮力判断每一个特殊字符的，如果是两个10、11，则第一位一定是1，让bit数组跳过一个字符即i=i+2

如果是一个字符0，则让bit数组继续遍历0后的下一个字符。直到判断到倒数第二个数字为止，如果这个数字是1，则返回false，如果这个数字是0，则返回true。

class Solution {
public:
    bool isOneBitCharacter(vector<int>& bits) {
        int n = bits.size(), flag = true;
        for (int i = 0; i != n;) {
            if (bits[i] == 1) {
                if (i == n - 2) {
                    flag = false;
                    break;
                }
                else
                    i = i + 2;
            }
            else {
                if (i == n - 2) {
                    flag = true;
                    break;
                }
                else
                    i = i + 1;
            }
        }
        return flag;
    }
};
// 3 ms