481. Magical String¶
Problem:¶
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.
Solutions:¶
public class Solution {
public int magicalString(int n) {
if (n <= 0) {
return 0;
}
int[] data = new int[100002];
data[0] = 1;
int i = 0, j = 0;
int count = 1;
while (i < n) {
if (data[j] == 1) {
if (data[i] == 1) {
data[++i] = 2;
}
else {
data[++i] = 1;
if (i < n) {
count ++;
}
}
}
else {
if (data[i] == 1) {
data[++i] = 1;
if (i < n) {
count ++;
}
data[++i] = 2;
}
else {
data[++i] = 2;
data[++i] = 1;
if (i < n) {
count ++;
}
}
}
j ++;
}
return count;
}
}
public class Solution {
public int magicalString(int n) {
if (n <= 0) {
return 0;
}
if (n <= 3) {
return 1;
}
int[] data = new int[100002];
data[0] = 1;
data[1] = 2;
data[2] = 2;
int i = 2, j = 2;
int count = 1;
int num = 1;
while (i < n) {
for (int k = 0; k < data[j]; k ++) {
data[++i] = num;
if (num == 1 && i < n) {
count ++;
}
}
j ++;
num = num ^ 3;
}
return count;
}
}
public class Solution {
public int magicalString(int n) {
if (n <= 0) {
return 0;
}
if (n <= 3) {
return 1;
}
StringBuilder sb = new StringBuilder("122");
int i = 2;
while (sb.length() < n) {
int num = (sb.charAt(sb.length() - 1) - '0') ^ 3;
int repeat = sb.charAt(i) - '0';
for (int j = 0; j < repeat; j ++) {
sb.append(num);
}
i ++;
}
int count = 0;
for (i = 0; i < n; i ++) {
if (sb.charAt(i) == '1') {
count ++;
}
}
return count;
}
}