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 s is 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 is s itself.
Given an integer n, return the number of 1's in the first n number in the magical string s.
Example 1:
Input: n = 6
Output: 3
Explanation: The first 6 elements of magical string s is "122112" and it contains three 1's, so return 3.
Example 2:
Input: n = 1
Output: 1
Constraints:
Solutions#
Solution 1#
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| class Solution:
def magicalString(self, n: int) -> int:
s = [1, 2, 2]
i = 2
while len(s) < n:
pre = s[-1]
cur = 3 - pre
# cur 表示这一组的数字，s[i] 表示这一组数字出现的次数
s += [cur] * s[i]
i += 1
return s[:n].count(1)
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| class Solution {
public int magicalString(int n) {
List<Integer> s = new ArrayList<>(Arrays.asList(1, 2, 2));
for (int i = 2; s.size() < n; ++i) {
int pre = s.get(s.size() - 1);
int cur = 3 - pre;
for (int j = 0; j < s.get(i); ++j) {
s.add(cur);
}
}
int ans = 0;
for (int i = 0; i < n; ++i) {
if (s.get(i) == 1) {
++ans;
}
}
return ans;
}
}
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| class Solution {
public:
int magicalString(int n) {
vector<int> s = {1, 2, 2};
for (int i = 2; s.size() < n; ++i) {
int pre = s.back();
int cur = 3 - pre;
for (int j = 0; j < s[i]; ++j) {
s.emplace_back(cur);
}
}
return count(s.begin(), s.begin() + n, 1);
}
};
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| func magicalString(n int) (ans int) {
s := []int{1, 2, 2}
for i := 2; len(s) < n; i++ {
pre := s[len(s)-1]
cur := 3 - pre
for j := 0; j < s[i]; j++ {
s = append(s, cur)
}
}
for _, c := range s[:n] {
if c == 1 {
ans++
}
}
return
}
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| function magicalString(n: number): number {
const cs = [...'1221121'];
let i = 5;
while (cs.length < n) {
const c = cs[cs.length - 1];
cs.push(c === '1' ? '2' : '1');
if (cs[i] !== '1') {
cs.push(c === '1' ? '2' : '1');
}
i++;
}
return cs.slice(0, n).reduce((r, c) => r + (c === '1' ? 1 : 0), 0);
}
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| impl Solution {
pub fn magical_string(n: i32) -> i32 {
let n = n as usize;
let mut s = String::from("1221121");
let mut i = 5;
while s.len() < n {
let c = s.as_bytes()[s.len() - 1];
s.push(if c == b'1' { '2' } else { '1' });
if s.as_bytes()[i] != b'1' {
s.push(if c == b'1' { '2' } else { '1' });
}
i += 1;
}
s
.as_bytes()
[0..n].iter()
.filter(|&v| v == &b'1')
.count() as i32
}
}
|
