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feat: add solutions to lcof2 problem: No.016 (doocs#1470)
No.016.不含重复字符的最长子字符串
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‎lcof2/剑指 Offer II 016. 不含重复字符的最长子字符串/README.md

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@@ -57,7 +57,13 @@
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<!-- 这里可写通用的实现逻辑 -->
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因为 `s` 中可能会出现字母、数字、符号和空格，所以可以用哈希表表示窗口
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**方法一：双指针 + 哈希表**
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我们用两个指针 $j$ 和 $i$ 维护一个不包含重复字符的子串，其中 $j$ 为子串的左边界，$i$ 为子串的右边界，用一个哈希表或数组 $ss$ 记录窗口中所有出现过的字符。
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接下来，我们遍历字符串 $s$，对于当前遍历到的字符 $s[i]$，如果 $s[i]$ 在 $[j, i)$ 范围内有与 $s[i]$ 相同的字符，我们就不断地向右移动指针 $j$，直到 $ss[s[i]]$ 为 `false`，此时 $[j,i)$ 中没有任何与 $s[i]$ 相同的字符，我们就找到了以字符 $s[i]$ 为结尾的最长子串。对于每个 $i$，我们都更新最长子串的长度，最终返回答案。
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时间复杂度 $O(n)$，空间复杂度 $O(|\Sigma|)$，其中 $n$ 为字符串 $s$ 的长度，而 $\Sigma$ 表示字符集，本题中字符集为所有 ASCII 码在 $[0, 128)$ 内的字符，即 $|\Sigma|=128$。
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<!-- tabs:start -->
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@@ -68,17 +74,14 @@
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```python
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class Solution:
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def lengthOfLongestSubstring(self, s: str) -> int:
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window = defaultdict(int)
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n, ans = len(s), 0
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left, right = 0, 0
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while right < n:
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ch = s[right]
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right += 1
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window[ch] += 1
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while window[ch] > 1:
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window[s[left]] -= 1
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left += 1
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ans = max(ans, right - left)
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ss = set()
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ans = j = 0
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for i, c in enumerate(s):
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while c in ss:
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ss.remove(s[j])
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j += 1
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ans = max(ans, i - j + 1)
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ss.add(c)
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return ans
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```
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@@ -89,41 +92,58 @@ class Solution:
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```java
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class Solution {
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public int lengthOfLongestSubstring(String s) {
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Map<Character, Integer> window = new HashMap<>();
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int n = s.length(), ans = 0;
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int left = 0, right = 0;
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while (right < n) {
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char ch = s.charAt(right++);
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window.merge(ch, 1, Integer::sum);
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while (window.get(ch) > 1) {
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window.merge(s.charAt(left++), -1, Integer::sum);
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boolean[] ss = new boolean[128];
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int ans = 0, j = 0;
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int n = s.length();
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for (int i = 0; i < n; ++i) {
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char c = s.charAt(i);
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while (ss[c]) {
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ss[s.charAt(j++)] = false;
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}
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ans = Math.max(ans, right - left);
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ans = Math.max(ans, i - j + 1);
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ss[c] = true;
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int lengthOfLongestSubstring(string s) {
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bool ss[128] = {false};
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int n = s.size();
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int ans = 0;
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for (int i = 0, j = 0; i < n; ++i) {
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while (ss[s[i]]) {
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ss[s[j++]] = false;
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}
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ss[s[i]] = true;
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ans = max(ans, i - j + 1);
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func lengthOfLongestSubstring(s string) int {
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window := make(map[byte]int)
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n := len(s)
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ans := 0
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left, right := 0, 0
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for right < n {
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ch := s[right]
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right++
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window[ch]++
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for window[ch] > 1 {
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window[s[left]]--
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left++
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func lengthOfLongestSubstring(s string) (ans int) {
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ss := make([]bool, 128)
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j := 0
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for i, c := range s {
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for ss[c] {
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ss[s[j]] = false
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j++
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}
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ans = max(ans, right-left)
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ss[c] = true
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ans = max(ans, i-j+1)
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}
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return ans
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return
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}
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func max(a, b int) int {
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int lengthOfLongestSubstring(string s) {
143-
if (s.size() == 0)
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return 0;
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int left = 0;
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int maxlen = 0;
148-
unordered_set<char> hash;
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for (int right = 0; right < s.size(); right++) {
151-
while (hash.find(s[right]) != hash.end()) {
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hash.erase(s[left]);
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left++;
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}
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### **TypeScript**
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156-
hash.insert(s[right]);
157-
maxlen = max(maxlen, right - left + 1);
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```ts
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function lengthOfLongestSubstring(s: string): number {
161+
let ans = 0;
162+
const n = s.length;
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const ss: boolean[] = new Array(128).fill(false);
164+
for (let i = 0, j = 0; i < n; ++i) {
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while (ss[s.charCodeAt(i)]) {
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ss[s.charCodeAt(j++)] = false;
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}
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160-
return maxlen;
168+
ss[s.charCodeAt(i)] = true;
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ans = Math.max(ans, i - j + 1);
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}
162-
};
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return ans;
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}
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```
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### **...**

‎lcof2/剑指 Offer II 016. 不含重复字符的最长子字符串/Solution.cpp

Lines changed: 9 additions & 16 deletions
Original file line numberDiff line numberDiff line change
@@ -1,23 +1,16 @@
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class Solution {
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public:
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int lengthOfLongestSubstring(string s) {
4-
if (s.size() == 0)
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return 0;
6-
7-
int left = 0;
8-
int maxlen = 0;
9-
unordered_set<char> hash;
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11-
for (int right = 0; right < s.size(); right++) {
12-
while (hash.find(s[right]) != hash.end()) {
13-
hash.erase(s[left]);
14-
left++;
4+
bool ss[128] = {false};
5+
int n = s.size();
6+
int ans = 0;
7+
for (int i = 0, j = 0; i < n; ++i) {
8+
while (ss[s[i]]) {
9+
ss[s[j++]] = false;
1510
}
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17-
hash.insert(s[right]);
18-
maxlen = max(maxlen, right - left + 1);
11+
ss[s[i]] = true;
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ans = max(ans, i - j + 1);
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}
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21-
return maxlen;
14+
return ans;
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}
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};

‎lcof2/剑指 Offer II 016. 不含重复字符的最长子字符串/Solution.go

Lines changed: 10 additions & 14 deletions
Original file line numberDiff line numberDiff line change
@@ -1,19 +1,15 @@
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func lengthOfLongestSubstring(s string) int {
2-
window := make(map[byte]int)
3-
n := len(s)
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ans := 0
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left, right := 0, 0
6-
for right < n {
7-
ch := s[right]
8-
right++
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window[ch]++
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for window[ch] > 1 {
11-
window[s[left]]--
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left++
1+
func lengthOfLongestSubstring(s string) (ans int) {
2+
ss := make([]bool, 128)
3+
j := 0
4+
for i, c := range s {
5+
for ss[c] {
6+
ss[s[j]] = false
7+
j++
138
}
14-
ans = max(ans, right-left)
9+
ss[c] = true
10+
ans = max(ans, i-j+1)
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}
16-
return ans
12+
return
1713
}
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1915
func max(a, b int) int {

‎lcof2/剑指 Offer II 016. 不含重复字符的最长子字符串/Solution.java

Lines changed: 16 additions & 16 deletions
Original file line numberDiff line numberDiff line change
@@ -1,16 +1,16 @@
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class Solution {
2-
public int lengthOfLongestSubstring(String s) {
3-
Map<Character, Integer> window = new HashMap<>();
4-
int n = s.length(), ans = 0;
5-
int left = 0, right = 0;
6-
while (right < n) {
7-
char ch = s.charAt(right++);
8-
window.merge(ch, 1, Integer::sum);
9-
while (window.get(ch) > 1) {
10-
window.merge(s.charAt(left++), -1, Integer::sum);
11-
}
12-
ans = Math.max(ans, right - left);
13-
}
14-
return ans;
15-
}
16-
}
1+
class Solution {
2+
public int lengthOfLongestSubstring(String s) {
3+
boolean[] ss = new boolean[128];
4+
int ans = 0, j = 0;
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int n = s.length();
6+
for (int i = 0; i < n; ++i) {
7+
char c = s.charAt(i);
8+
while (ss[c]) {
9+
ss[s.charAt(j++)] = false;
10+
}
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ans = Math.max(ans, i - j + 1);
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ss[c] = true;
13+
}
14+
return ans;
15+
}
16+
}

‎lcof2/剑指 Offer II 016. 不含重复字符的最长子字符串/Solution.py

Lines changed: 11 additions & 14 deletions
Original file line numberDiff line numberDiff line change
@@ -1,14 +1,11 @@
1-
class Solution:
2-
def lengthOfLongestSubstring(self, s: str) -> int:
3-
window = defaultdict(int)
4-
n, ans = len(s), 0
5-
left, right = 0, 0
6-
while right < n:
7-
ch = s[right]
8-
right += 1
9-
window[ch] += 1
10-
while window[ch] > 1:
11-
window[s[left]] -= 1
12-
left += 1
13-
ans = max(ans, right - left)
14-
return ans
1+
class Solution:
2+
def lengthOfLongestSubstring(self, s: str) -> int:
3+
ss = set()
4+
ans = j = 0
5+
for i, c in enumerate(s):
6+
while c in ss:
7+
ss.remove(s[j])
8+
j += 1
9+
ans = max(ans, i - j + 1)
10+
ss.add(c)
11+
return ans

‎lcof2/剑指 Offer II 016. 不含重复字符的最长子字符串/Solution.ts

Lines changed: 13 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,13 @@
1+
function lengthOfLongestSubstring(s: string): number {
2+
let ans = 0;
3+
const n = s.length;
4+
const ss: boolean[] = new Array(128).fill(false);
5+
for (let i = 0, j = 0; i < n; ++i) {
6+
while (ss[s.charCodeAt(i)]) {
7+
ss[s.charCodeAt(j++)] = false;
8+
}
9+
ss[s.charCodeAt(i)] = true;
10+
ans = Math.max(ans, i - j + 1);
11+
}
12+
return ans;
13+
}

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