First blog!
Let's start with this short interesting one.
Problem:
Find the max length of consecutive ones in a "01" vector.
Thinking process:
1. 0s divide 1s. So ++count of 1s while num == 1, reset it to 0 while num == 0. Update the ans to max each time.
2. To decrease comparison times: If 1s are more frequent than 0s, we can update ans whenever we meet a 0; but need one more update after traversing the vector.
3. On the contrary, If 0s are more frequent, update ans while num == 1.
4. Corner case: vector is empty => 0
Let's start with this short interesting one.
Problem:
Find the max length of consecutive ones in a "01" vector.
Thinking process:
1. 0s divide 1s. So ++count of 1s while num == 1, reset it to 0 while num == 0. Update the ans to max each time.
2. To decrease comparison times: If 1s are more frequent than 0s, we can update ans whenever we meet a 0; but need one more update after traversing the vector.
3. On the contrary, If 0s are more frequent, update ans while num == 1.
4. Corner case: vector is empty => 0
No comments:
Post a Comment