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Formatted question description: https://leetcode.ca/all/717.html
717. 1bit and 2bit Characters (Easy)
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 onebit 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 twobit character and onebit character. So the last character is onebit character.
Example 2:
Input: bits = [1, 1, 1, 0] Output: False Explanation: The only way to decode it is twobit character and twobit character. So the last character is NOT onebit character.
Note:
1 <= len(bits) <= 1000
.bits[i]
is always 0
or 1
.Related Topics:
Array
Similar Questions:
Solution 1.
// OJ: https://leetcode.com/problems/1bitand2bitcharacters/
// 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;
}
};

class Solution { public boolean isOneBitCharacter(int[] bits) { int length = bits.length; int index = 0; while (index < length) { if (index == length  1) return true; int bit = bits[index]; if (bit == 0) index++; else if (bit == 1) index += 2; } return false; } }

// OJ: https://leetcode.com/problems/1bitand2bitcharacters/ // 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; } };

class Solution: def isOneBitCharacter(self, bits: List[int]) > bool: i, n = 0, len(bits) while i < n  1: i += bits[i] + 1 return i == n  1 ############ """ class Solution(object): def isOneBitCharacter(self, bits): """ :type bits: List[int] :rtype: bool """ pos = 0 while pos < len(bits)  1: if bits[pos] == 1: pos += 2 else: pos += 1 return pos == len(bits)  1 and bits[pos] == 0