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Commit 894ebe6

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feat: add solutions to lc problems: No.0481,0482 (#3246)
* No.0481.Magical String * No.0482.License Key Formatting
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13 files changed

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-218
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13 files changed

+367
-218
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‎solution/0400-0499/0481.Magical String/README.md‎

Lines changed: 18 additions & 19 deletions
Original file line numberDiff line numberDiff line change
@@ -115,7 +115,7 @@ class Solution:
115115
```java
116116
class Solution {
117117
public int magicalString(int n) {
118-
List<Integer> s = new ArrayList<>(Arrays.asList(1, 2, 2));
118+
List<Integer> s = new ArrayList<>(List.of(1, 2, 2));
119119
for (int i = 2; s.size() < n; ++i) {
120120
int pre = s.get(s.size() - 1);
121121
int cur = 3 - pre;
@@ -178,17 +178,15 @@ func magicalString(n int) (ans int) {
178178

179179
```ts
180180
function magicalString(n: number): number {
181-
const cs = [...'1221121'];
182-
let i = 5;
183-
while (cs.length < n) {
184-
const c = cs[cs.length - 1];
185-
cs.push(c === '1' ? '2' : '1');
186-
if (cs[i] !== '1') {
187-
cs.push(c === '1' ? '2' : '1');
181+
const s: number[] = [1, 2, 2];
182+
for (let i = 2; s.length < n; ++i) {
183+
let pre = s[s.length - 1];
184+
let cur = 3 - pre;
185+
for (let j = 0; j < s[i]; ++j) {
186+
s.push(cur);
188187
}
189-
i++;
190188
}
191-
return cs.slice(0, n).reduce((r, c) => r+ (c=== '1'?1:0), 0);
189+
return s.slice(0, n).filter(x=> x=== 1).length;
192190
}
193191
```
194192

@@ -197,18 +195,19 @@ function magicalString(n: number): number {
197195
```rust
198196
impl Solution {
199197
pub fn magical_string(n: i32) -> i32 {
200-
let n=nasusize;
201-
let mut s = String::from("1221121");
202-
letmuti=5;
203-
while s.len() < n {
204-
let c = s.as_bytes()[s.len() - 1];
205-
s.push(ifc==b'1' { '2' } else { '1' });
206-
ifs.as_bytes()[i]!=b'1' {
207-
s.push(ifc==b'1' { '2' } else { '1' });
198+
let muts=vec![1, 2, 2];
199+
let mut i = 2;
200+
201+
while s.len() < n asusize{
202+
let pre = s[s.len() - 1];
203+
letcur=3-pre;
204+
for_in0..s[i] {
205+
s.push(cur);
208206
}
209207
i += 1;
210208
}
211-
s.as_bytes()[0..n].iter().filter(|&v| v == &b'1').count() as i32
209+
210+
s.iter().take(n as usize).filter(|&&x| x == 1).count() as i32
212211
}
213212
}
214213
```

‎solution/0400-0499/0481.Magical String/README_EN.md‎

Lines changed: 51 additions & 21 deletions
Original file line numberDiff line numberDiff line change
@@ -56,7 +56,39 @@ tags:
5656

5757
<!-- solution:start -->
5858

59-
### Solution 1
59+
### Solution 1: Simulate the Construction Process
60+
61+
According to the problem, we know that each group of numbers in the string $s$ can be obtained from the digits of the string $s$ itself.
62+
63+
The first two groups of numbers in string $s$ are 1ドル$ and 22ドル,ドル which are obtained from the first and second digits of string $s,ドル respectively. Moreover, the first group of numbers contains only 1ドル,ドル the second group contains only 2ドル,ドル the third group contains only 1ドル,ドル and so on.
64+
65+
Since the first two groups of numbers are known, we initialize string $s$ as 122ドル,ドル and then start constructing from the third group. The third group of numbers is obtained from the third digit of string $s$ (index $i=2$), so at this point, we point the pointer $i$ to the third digit 2ドル$ of string $s$.
66+
67+
```
68+
1 2 2
69+
^
70+
i
71+
```
72+
73+
The digit pointed by pointer $i$ is 2ドル,ドル indicating that the third group of numbers will appear twice. Since the previous group of numbers is 2ドル,ドル and the numbers alternate between groups, the third group of numbers is two 1ドル$s, i.e., 11ドル$. After construction, the pointer $i$ moves to the next position, pointing to the fourth digit 1ドル$ of string $s$.
74+
75+
```
76+
1 2 2 1 1
77+
^
78+
i
79+
```
80+
81+
At this point, the digit pointed by pointer $i$ is 1ドル,ドル indicating that the fourth group of numbers will appear once. Since the previous group of numbers is 1ドル,ドル and the numbers alternate between groups, the fourth group of numbers is one 2ドル,ドル i.e., 2ドル$. After construction, the pointer $i$ moves to the next position, pointing to the fifth digit 1ドル$ of string $s$.
82+
83+
```
84+
1 2 2 1 1 2
85+
^
86+
i
87+
```
88+
89+
Following this rule, we simulate the construction process sequentially until the length of string $s$ is greater than or equal to $n$.
90+
91+
The time complexity is $O(n),ドル and the space complexity is $O(n)$.
6092

6193
<!-- tabs:start -->
6294

@@ -70,7 +102,6 @@ class Solution:
70102
while len(s) < n:
71103
pre = s[-1]
72104
cur = 3 - pre
73-
# cur 表示这一组的数字,s[i] 表示这一组数字出现的次数
74105
s += [cur] * s[i]
75106
i += 1
76107
return s[:n].count(1)
@@ -81,7 +112,7 @@ class Solution:
81112
```java
82113
class Solution {
83114
public int magicalString(int n) {
84-
List<Integer> s = new ArrayList<>(Arrays.asList(1, 2, 2));
115+
List<Integer> s = new ArrayList<>(List.of(1, 2, 2));
85116
for (int i = 2; s.size() < n; ++i) {
86117
int pre = s.get(s.size() - 1);
87118
int cur = 3 - pre;
@@ -144,17 +175,15 @@ func magicalString(n int) (ans int) {
144175

145176
```ts
146177
function magicalString(n: number): number {
147-
const cs = [...'1221121'];
148-
let i = 5;
149-
while (cs.length < n) {
150-
const c = cs[cs.length - 1];
151-
cs.push(c === '1' ? '2' : '1');
152-
if (cs[i] !== '1') {
153-
cs.push(c === '1' ? '2' : '1');
178+
const s: number[] = [1, 2, 2];
179+
for (let i = 2; s.length < n; ++i) {
180+
let pre = s[s.length - 1];
181+
let cur = 3 - pre;
182+
for (let j = 0; j < s[i]; ++j) {
183+
s.push(cur);
154184
}
155-
i++;
156185
}
157-
return cs.slice(0, n).reduce((r, c) => r+ (c=== '1'?1:0), 0);
186+
return s.slice(0, n).filter(x=> x=== 1).length;
158187
}
159188
```
160189

@@ -163,18 +192,19 @@ function magicalString(n: number): number {
163192
```rust
164193
impl Solution {
165194
pub fn magical_string(n: i32) -> i32 {
166-
let n=nasusize;
167-
let mut s = String::from("1221121");
168-
letmuti=5;
169-
while s.len() < n {
170-
let c = s.as_bytes()[s.len() - 1];
171-
s.push(ifc==b'1' { '2' } else { '1' });
172-
ifs.as_bytes()[i]!=b'1' {
173-
s.push(ifc==b'1' { '2' } else { '1' });
195+
let muts=vec![1, 2, 2];
196+
let mut i = 2;
197+
198+
while s.len() < n asusize{
199+
let pre = s[s.len() - 1];
200+
letcur=3-pre;
201+
for_in0..s[i] {
202+
s.push(cur);
174203
}
175204
i += 1;
176205
}
177-
s.as_bytes()[0..n].iter().filter(|&v| v == &b'1').count() as i32
206+
207+
s.iter().take(n as usize).filter(|&&x| x == 1).count() as i32
178208
}
179209
}
180210
```

‎solution/0400-0499/0481.Magical String/Solution.java‎

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -1,6 +1,6 @@
11
class Solution {
22
public int magicalString(int n) {
3-
List<Integer> s = new ArrayList<>(Arrays.asList(1, 2, 2));
3+
List<Integer> s = new ArrayList<>(List.of(1, 2, 2));
44
for (int i = 2; s.size() < n; ++i) {
55
int pre = s.get(s.size() - 1);
66
int cur = 3 - pre;

‎solution/0400-0499/0481.Magical String/Solution.py‎

Lines changed: 0 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -5,7 +5,6 @@ def magicalString(self, n: int) -> int:
55
while len(s) < n:
66
pre = s[-1]
77
cur = 3 - pre
8-
# cur 表示这一组的数字,s[i] 表示这一组数字出现的次数
98
s += [cur] * s[i]
109
i += 1
1110
return s[:n].count(1)
Lines changed: 10 additions & 9 deletions
Original file line numberDiff line numberDiff line change
@@ -1,16 +1,17 @@
11
impl Solution {
22
pub fn magical_string(n: i32) -> i32 {
3-
let n = n asusize;
4-
let mut s = String::from("1221121");
5-
letmut i = 5;
6-
while s.len() < n {
7-
let c = s.as_bytes()[s.len() - 1];
8-
s.push(if c == b'1'{'2'}else{'1'});
9-
if s.as_bytes()[i] != b'1' {
10-
s.push(if c == b'1'{'2'}else{'1'});
3+
let mut s = vec![1,2,2];
4+
let mut i = 2;
5+
6+
while s.len() < n asusize{
7+
let pre = s[s.len() - 1];
8+
let cur = 3 - pre;
9+
for _ in0..s[i] {
10+
s.push(cur);
1111
}
1212
i += 1;
1313
}
14-
s.as_bytes()[0..n].iter().filter(|&v| v == &b'1').count() as i32
14+
15+
s.iter().take(n as usize).filter(|&&x| x == 1).count() as i32
1516
}
1617
}
Lines changed: 7 additions & 9 deletions
Original file line numberDiff line numberDiff line change
@@ -1,13 +1,11 @@
11
function magicalString(n: number): number {
2-
const cs = [...'1221121'];
3-
let i = 5;
4-
while (cs.length < n) {
5-
const c = cs[cs.length - 1];
6-
cs.push(c === '1' ? '2' : '1');
7-
if (cs[i] !== '1') {
8-
cs.push(c === '1' ? '2' : '1');
2+
const s: number[] = [1, 2, 2];
3+
for (let i = 2; s.length < n; ++i) {
4+
let pre = s[s.length - 1];
5+
let cur = 3 - pre;
6+
for (let j = 0; j < s[i]; ++j) {
7+
s.push(cur);
98
}
10-
i++;
119
}
12-
return cs.slice(0, n).reduce((r,c)=> r+(c=== '1' ? 1 : 0),0);
10+
return s.slice(0, n).filter(x=> x=== 1).length;
1311
}

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