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.
直接判断就好了。对于当前位,是1的话直接条到i+2是0的话不用管,然后看最后一位是不是0
class Solution {
public:
bool isOneBitCharacter(vector<int>& bits) {
for (int i = 0; i < bits.size(); ++i) {
if (bits[i] == 1) {
if (i + 1 < bits.size()) {
if (bits[i+1] == 0 || bits[i+1] == 1) {
i++;continue;
}
else {
return false;
}
} else {
return false;
}
}
else {
if (i == bits.size()-1) return true;
}
}
return false;
}
};