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();
            ;;
            ){
                )
                    index +=;
                else
                    index++;
    
            }
            ;
        }
    };

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