Formatted question description: https://leetcode.ca/all/481.html
481. Magical String
Level
Medium
Description
A magical string S consists of only ‘1’ and ‘2’ and obeys the following rules:
The string S is magical because concatenating the number of contiguous occurrences of characters ‘1’ and ‘2’ generates the string S itself.
The first few elements of string S is the following: S = “1221121221221121122……”
If we group the consecutive ‘1’s and ‘2’s in S, it will be:
1 22 11 2 1 22 1 22 11 2 11 22 ……
and the occurrences of ‘1’s or ‘2’s in each group are:
1 2 2 1 1 2 1 2 2 1 2 2 ……
You can see that the occurrence sequence above is the S itself.
Given an integer N as input, return the number of ‘1’s in the first N number in the magical string S.
Note: N will not exceed 100,000.
Example 1:
Input: 6
Output: 3
Explanation: The first 6 elements of magical string S is “12211” and it contains three 1’s, so return 3.
Solution
Generate the string directly. Obviously, the first three characters are “122”. Read each character and generate a group using a different character than the character before, until the string’s length is at least n. Loop over the first n characters and count the number of 1’s.
class Solution {
public int magicalString(int n) {
StringBuffer sb = new StringBuffer();
int index = 0;
int digit = 1;
while (sb.length() < n) {
if (index >= sb.length()) {
for (int i = 0; i < digit; i++)
sb.append(digit);
} else {
int curDigit = sb.charAt(index) - '0';
for (int i = 0; i < curDigit; i++)
sb.append(digit);
}
digit = 3 - digit;
index++;
}
String str = sb.substring(0, n).toString();
int ones = 0;
for (int i = 0; i < n; i++) {
if (str.charAt(i) == '1')
ones++;
}
return ones;
}
}