Q:
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
.
A:
本以为 需要DP, 结果不用。
class Solution {
public:
bool isOneBitCharacter(vector<int>& bits) {
return helper(bits, 0);
}
private:
bool helper(vector<int> &bits, int i)
{
int n = bits.size();
if(i>=n)
return false;
if(i==n-1)
return bits[i]==0;
// we have at least two bits
if(bits[i]==0){
return helper(bits,i+1);
}else{
return helper(bits, i+2);
}
}
};
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