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Commit 0caf311

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feat: update solutions to lc problems: No.0340~0342 (doocs#3288)
1 parent 6a59c00 commit 0caf311

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18 files changed

+429
-606
lines changed

18 files changed

+429
-606
lines changed

‎solution/0300-0399/0340.Longest Substring with At Most K Distinct Characters/README.md‎

Lines changed: 60 additions & 47 deletions
Original file line numberDiff line numberDiff line change
@@ -54,11 +54,13 @@ tags:
5454

5555
### 方法一:滑动窗口 + 哈希表
5656

57-
我们可以使用滑动窗口的思想,维护一个滑动窗口,使得窗口内的字符串中不同字符的个数不超过 $k$ 个。窗口内不同字符个数的统计可以用哈希表 `cnt` 来维护。
57+
我们可以使用滑动窗口的思想,用一个哈希表 $\textit{cnt}$ 记录窗口中每个字符的出现次数,用 $\textit{l}$ 记录窗口的左边界。
5858

59-
我们使用两个指针 $j$ 和 $i$ 分别表示滑动窗口的左右边界。我们先移动右边界 $i,ドル将字符 $s[i]$ 加入到窗口内,扩大滑动窗口,若此时窗口内不同字符的个数超过 $k$ 个,则移动左边界 $j,ドル缩小滑动窗口,直到窗口内不同字符的个数不超过 $k$ 个。此时我们可以更新答案的最大值,即 $ans = max(ans, i - j + 1)$。
59+
遍历字符串,每次将右边界的字符加入哈希表,如果哈希表中不同字符的个数超过了 $k$,则将左边界的字符从哈希表中删除,然后更新左边界 $\textit{l}$。
6060

61-
时间复杂度 $O(n),ドル空间复杂度 $O(\min(n, k))$。其中 $n$ 为字符串的长度。
61+
最后返回字符串的长度减去左边界的长度即可。
62+
63+
时间复杂度 $O(n),ドル空间复杂度 $O(k)$。其中 $n$ 为字符串的长度。
6264

6365
<!-- tabs:start -->
6466

@@ -67,18 +69,16 @@ tags:
6769
```python
6870
class Solution:
6971
def lengthOfLongestSubstringKDistinct(self, s: str, k: int) -> int:
72+
l = 0
7073
cnt = Counter()
71-
n = len(s)
72-
ans = j = 0
73-
for i, c in enumerate(s):
74+
for c in s:
7475
cnt[c] += 1
75-
while len(cnt) > k:
76-
cnt[s[j]] -= 1
77-
if cnt[s[j]] == 0:
78-
cnt.pop(s[j])
79-
j += 1
80-
ans = max(ans, i - j + 1)
81-
return ans
76+
if len(cnt) > k:
77+
cnt[s[l]] -= 1
78+
if cnt[s[l]] == 0:
79+
del cnt[s[l]]
80+
l += 1
81+
return len(s) - l
8282
```
8383

8484
#### Java
@@ -87,22 +87,18 @@ class Solution:
8787
class Solution {
8888
public int lengthOfLongestSubstringKDistinct(String s, int k) {
8989
Map<Character, Integer> cnt = new HashMap<>();
90-
int n = s.length();
91-
int ans = 0, j = 0;
92-
for (int i = 0; i < n; ++i) {
93-
char c = s.charAt(i);
94-
cnt.put(c, cnt.getOrDefault(c, 0) + 1);
95-
while (cnt.size() > k) {
96-
char t = s.charAt(j);
97-
cnt.put(t, cnt.getOrDefault(t, 0) - 1);
98-
if (cnt.get(t) == 0) {
99-
cnt.remove(t);
90+
int l = 0;
91+
char[] cs = s.toCharArray();
92+
for (char c : cs) {
93+
cnt.merge(c, 1, Integer::sum);
94+
if (cnt.size() > k) {
95+
if (cnt.merge(cs[l], -1, Integer::sum) == 0) {
96+
cnt.remove(cs[l]);
10097
}
101-
++j;
98+
++l;
10299
}
103-
ans = Math.max(ans, i - j + 1);
104100
}
105-
return ans;
101+
return cs.length - l;
106102
}
107103
}
108104
```
@@ -114,41 +110,58 @@ class Solution {
114110
public:
115111
int lengthOfLongestSubstringKDistinct(string s, int k) {
116112
unordered_map<char, int> cnt;
117-
int n = s.size();
118-
int ans = 0, j = 0;
119-
for (int i = 0; i < n; ++i) {
120-
cnt[s[i]]++;
121-
while (cnt.size() > k) {
122-
if (--cnt[s[j]] == 0) {
123-
cnt.erase(s[j]);
113+
int l = 0;
114+
for (char& c : s) {
115+
++cnt[c];
116+
if (cnt.size() > k) {
117+
if (--cnt[s[l]] == 0) {
118+
cnt.erase(s[l]);
124119
}
125-
++j;
120+
++l;
126121
}
127-
ans = max(ans, i - j + 1);
128122
}
129-
return ans;
123+
return s.size() - l;
130124
}
131125
};
132126
```
133127
134128
#### Go
135129
136130
```go
137-
func lengthOfLongestSubstringKDistinct(s string, k int) (ans int) {
131+
func lengthOfLongestSubstringKDistinct(s string, k int) int {
138132
cnt := map[byte]int{}
139-
j := 0
140-
for i := range s {
141-
cnt[s[i]]++
142-
for len(cnt) > k {
143-
cnt[s[j]]--
144-
if cnt[s[j]] == 0 {
145-
delete(cnt, s[j])
133+
l := 0
134+
for _, c := range s {
135+
cnt[byte(c)]++
136+
if len(cnt) > k {
137+
cnt[s[l]]--
138+
if cnt[s[l]] == 0 {
139+
delete(cnt, s[l])
146140
}
147-
j++
141+
l++
148142
}
149-
ans = max(ans, i-j+1)
150143
}
151-
return
144+
return len(s) - l
145+
}
146+
```
147+
148+
#### TypeScript
149+
150+
```ts
151+
function lengthOfLongestSubstringKDistinct(s: string, k: number): number {
152+
const cnt: Map<string, number> = new Map();
153+
let l = 0;
154+
for (const c of s) {
155+
cnt.set(c, (cnt.get(c) ?? 0) + 1);
156+
if (cnt.size > k) {
157+
cnt.set(s[l], cnt.get(s[l])! - 1);
158+
if (cnt.get(s[l]) === 0) {
159+
cnt.delete(s[l]);
160+
}
161+
l++;
162+
}
163+
}
164+
return s.length - l;
152165
}
153166
```
154167

‎solution/0300-0399/0340.Longest Substring with At Most K Distinct Characters/README_EN.md‎

Lines changed: 64 additions & 45 deletions
Original file line numberDiff line numberDiff line change
@@ -50,7 +50,15 @@ tags:
5050

5151
<!-- solution:start -->
5252

53-
### Solution 1
53+
### Solution 1: Sliding Window + Hash Table
54+
55+
We can use the idea of a sliding window, with a hash table $\textit{cnt}$ to record the occurrence count of each character within the window, and $\textit{l}$ to denote the left boundary of the window.
56+
57+
Iterate through the string, adding the character at the right boundary to the hash table each time. If the number of distinct characters in the hash table exceeds $k,ドル remove the character at the left boundary from the hash table, then update the left boundary $\textit{l}$.
58+
59+
Finally, return the length of the string minus the length of the left boundary.
60+
61+
The time complexity is $O(n),ドル and the space complexity is $O(k)$. Here, $n$ is the length of the string.
5462

5563
<!-- tabs:start -->
5664

@@ -59,18 +67,16 @@ tags:
5967
```python
6068
class Solution:
6169
def lengthOfLongestSubstringKDistinct(self, s: str, k: int) -> int:
70+
l = 0
6271
cnt = Counter()
63-
n = len(s)
64-
ans = j = 0
65-
for i, c in enumerate(s):
72+
for c in s:
6673
cnt[c] += 1
67-
while len(cnt) > k:
68-
cnt[s[j]] -= 1
69-
if cnt[s[j]] == 0:
70-
cnt.pop(s[j])
71-
j += 1
72-
ans = max(ans, i - j + 1)
73-
return ans
74+
if len(cnt) > k:
75+
cnt[s[l]] -= 1
76+
if cnt[s[l]] == 0:
77+
del cnt[s[l]]
78+
l += 1
79+
return len(s) - l
7480
```
7581

7682
#### Java
@@ -79,22 +85,18 @@ class Solution:
7985
class Solution {
8086
public int lengthOfLongestSubstringKDistinct(String s, int k) {
8187
Map<Character, Integer> cnt = new HashMap<>();
82-
int n = s.length();
83-
int ans = 0, j = 0;
84-
for (int i = 0; i < n; ++i) {
85-
char c = s.charAt(i);
86-
cnt.put(c, cnt.getOrDefault(c, 0) + 1);
87-
while (cnt.size() > k) {
88-
char t = s.charAt(j);
89-
cnt.put(t, cnt.getOrDefault(t, 0) - 1);
90-
if (cnt.get(t) == 0) {
91-
cnt.remove(t);
88+
int l = 0;
89+
char[] cs = s.toCharArray();
90+
for (char c : cs) {
91+
cnt.merge(c, 1, Integer::sum);
92+
if (cnt.size() > k) {
93+
if (cnt.merge(cs[l], -1, Integer::sum) == 0) {
94+
cnt.remove(cs[l]);
9295
}
93-
++j;
96+
++l;
9497
}
95-
ans = Math.max(ans, i - j + 1);
9698
}
97-
return ans;
99+
return cs.length - l;
98100
}
99101
}
100102
```
@@ -106,41 +108,58 @@ class Solution {
106108
public:
107109
int lengthOfLongestSubstringKDistinct(string s, int k) {
108110
unordered_map<char, int> cnt;
109-
int n = s.size();
110-
int ans = 0, j = 0;
111-
for (int i = 0; i < n; ++i) {
112-
cnt[s[i]]++;
113-
while (cnt.size() > k) {
114-
if (--cnt[s[j]] == 0) {
115-
cnt.erase(s[j]);
111+
int l = 0;
112+
for (char& c : s) {
113+
++cnt[c];
114+
if (cnt.size() > k) {
115+
if (--cnt[s[l]] == 0) {
116+
cnt.erase(s[l]);
116117
}
117-
++j;
118+
++l;
118119
}
119-
ans = max(ans, i - j + 1);
120120
}
121-
return ans;
121+
return s.size() - l;
122122
}
123123
};
124124
```
125125
126126
#### Go
127127
128128
```go
129-
func lengthOfLongestSubstringKDistinct(s string, k int) (ans int) {
129+
func lengthOfLongestSubstringKDistinct(s string, k int) int {
130130
cnt := map[byte]int{}
131-
j := 0
132-
for i := range s {
133-
cnt[s[i]]++
134-
for len(cnt) > k {
135-
cnt[s[j]]--
136-
if cnt[s[j]] == 0 {
137-
delete(cnt, s[j])
131+
l := 0
132+
for _, c := range s {
133+
cnt[byte(c)]++
134+
if len(cnt) > k {
135+
cnt[s[l]]--
136+
if cnt[s[l]] == 0 {
137+
delete(cnt, s[l])
138138
}
139-
j++
139+
l++
140140
}
141-
ans = max(ans, i-j+1)
142141
}
143-
return
142+
return len(s) - l
143+
}
144+
```
145+
146+
#### TypeScript
147+
148+
```ts
149+
function lengthOfLongestSubstringKDistinct(s: string, k: number): number {
150+
const cnt: Map<string, number> = new Map();
151+
let l = 0;
152+
for (const c of s) {
153+
cnt.set(c, (cnt.get(c) ?? 0) + 1);
154+
if (cnt.size > k) {
155+
cnt.set(s[l], cnt.get(s[l])! - 1);
156+
if (cnt.get(s[l]) === 0) {
157+
cnt.delete(s[l]);
158+
}
159+
l++;
160+
}
161+
}
162+
return s.length - l;
144163
}
145164
```
146165

‎solution/0300-0399/0340.Longest Substring with At Most K Distinct Characters/Solution.cpp‎

Lines changed: 8 additions & 10 deletions
Original file line numberDiff line numberDiff line change
@@ -2,18 +2,16 @@ class Solution {
22
public:
33
int lengthOfLongestSubstringKDistinct(string s, int k) {
44
unordered_map<char, int> cnt;
5-
int n = s.size();
6-
int ans = 0, j = 0;
7-
for (int i = 0; i < n; ++i) {
8-
cnt[s[i]]++;
9-
while (cnt.size() > k) {
10-
if (--cnt[s[j]] == 0) {
11-
cnt.erase(s[j]);
5+
int l = 0;
6+
for (char& c : s) {
7+
++cnt[c];
8+
if (cnt.size() > k) {
9+
if (--cnt[s[l]] == 0) {
10+
cnt.erase(s[l]);
1211
}
13-
++j;
12+
++l;
1413
}
15-
ans = max(ans, i - j + 1);
1614
}
17-
return ans;
15+
return s.size() - l;
1816
}
1917
};
Lines changed: 10 additions & 11 deletions
Original file line numberDiff line numberDiff line change
@@ -1,16 +1,15 @@
1-
func lengthOfLongestSubstringKDistinct(s string, k int) (ansint) {
1+
func lengthOfLongestSubstringKDistinct(s string, k int) int {
22
cnt := map[byte]int{}
3-
j := 0
4-
for i := range s {
5-
cnt[s[i]]++
6-
for len(cnt) > k {
7-
cnt[s[j]]--
8-
if cnt[s[j]] == 0 {
9-
delete(cnt, s[j])
3+
l := 0
4+
for _, c := range s {
5+
cnt[byte(c)]++
6+
if len(cnt) > k {
7+
cnt[s[l]]--
8+
if cnt[s[l]] == 0 {
9+
delete(cnt, s[l])
1010
}
11-
j++
11+
l++
1212
}
13-
ans = max(ans, i-j+1)
1413
}
15-
return
14+
returnlen(s) -l
1615
}
Lines changed: 9 additions & 13 deletions
Original file line numberDiff line numberDiff line change
@@ -1,21 +1,17 @@
11
class Solution {
22
public int lengthOfLongestSubstringKDistinct(String s, int k) {
33
Map<Character, Integer> cnt = new HashMap<>();
4-
int n = s.length();
5-
int ans = 0, j = 0;
6-
for (int i = 0; i < n; ++i) {
7-
char c = s.charAt(i);
8-
cnt.put(c, cnt.getOrDefault(c, 0) + 1);
9-
while (cnt.size() > k) {
10-
char t = s.charAt(j);
11-
cnt.put(t, cnt.getOrDefault(t, 0) - 1);
12-
if (cnt.get(t) == 0) {
13-
cnt.remove(t);
4+
int l = 0;
5+
char[] cs = s.toCharArray();
6+
for (char c : cs) {
7+
cnt.merge(c, 1, Integer::sum);
8+
if (cnt.size() > k) {
9+
if (cnt.merge(cs[l], -1, Integer::sum) == 0) {
10+
cnt.remove(cs[l]);
1411
}
15-
++j;
12+
++l;
1613
}
17-
ans = Math.max(ans, i - j + 1);
1814
}
19-
return ans;
15+
return cs.length - l;
2016
}
2117
}

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