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.想法：遍历vector，遇到1，指针加2，遇到0，指针加1，判断指针和数组最后一位是否相等。
class Solution { public: bool isOneBitCharacter(vector<int>& bits) { int len = bits.size(); int index = 0;; while(index < len-1){ if(bits[index] == 1) index +=2; else index++; } return index == len -1; } };