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
.Related Topics:
Array
Similar Questions:
// OJ: https://leetcode.com/problems/1-bit-and-2-bit-characters/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
bool isOneBitCharacter(vector<int>& A) {
for (int i = 0, N = A.size(); i < N; ) {
if (A[i] == 1) {
if (i + 2 < N) i += 2;
else return false;
} else if (i == N - 1) return true;
else ++i;
}
return false;
}
};