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.
题目大意：我们有两个特殊字符。 第一个字符可以用1位表示。第二个字符可以用2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是一位字符。 给定的字符串将始终以0结束
分析：i从0开始遍历整个bits数组，当i遇到0时走一步，否则走2步，判断是否会走到最后一个元素～
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class Solution { public: bool isOneBitCharacter(vector<int>& bits) { int i = 0; while(i < bits.size()) { if (i == (bits.size() - 1)) return true; if (bits[i] == 0) i++; else i += 2; } return false; } }; |
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