problem:
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 always0or1.
solution:
主要题意就是当一位数为1的时候,后面必须跟一个数,然后10或者11组成一个two-character的东西,然后判断0是不是在这个东西里面。
class Solution {
public:
bool isOneBitCharacter(vector<int>& bits) {
int flag = 0;
for (int i = 0; i < bits.size() - 1; i++)
{
if (bits[i] == 1)
{
i++;
}
if (i == bits.size()-1)
{
return false;
}
}
return true;
}
};
本文探讨了一种特殊字符编码方案的问题,其中一种字符由一个比特位表示,另一种由两个比特位表示。文章通过示例解释了如何确定字符串是否以单比特字符结尾,并提供了解决方案的代码实现。

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