1-bit and 2-bit Characters(leetcode)

1-bit and 2-bit Characters

1-bit and 2-bit Characters
题目

解决

题目

leetcode题目

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

.

解决

遍历给定的字符串：

若遇到

bits[i] == 0

，跳过，不做任何处理

若遇到

bits[i] == 1

，判断是否处在最后一位(无法构建题目中的character)或者倒数第二位(自动与最后一位的

bits[n - 1]

构成character)。若满足

i == n - 1 || i + 1 == n - 1

，都不能使得最后的

bit

变成一位的character。若不满足，接着判断

bits[i + 2]

，直到循环结束。

class Solution {
public:
bool isOneBitCharacter(vector<int>& bits) {
int num = bits.size();
for (int i = 0; i < num; i++) {
if (bits[i] == 1) {
if (i == num - 1 || i + 1 == num - 1) {
return false;
}
i++;
}
}
return true;
}
};