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Commit 11b6e20

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feat: add solutions to leetcode problem: No.0438. Find All Anagrams in a String
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‎solution/0400-0499/0438.Find All Anagrams in a String/README.md‎

Lines changed: 72 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -46,27 +46,97 @@ s: "abab" p: "ab"
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起始索引等于 2 的子串是 "ab", 它是 "ab" 的字母异位词。
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</pre>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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"双指针 + 滑动窗口"求解。定义滑动窗口的左右两个指针 left、right,right 一步步往右走遍历 s 字符串,当 right 指针遍历到的字符加入 t 后超过 counter 的字符数量时,将滑动窗口左侧字符不断弹出,也就是 left 指针不断右移,直到符合要求为止。
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若滑动窗口长度等于字符串 p 的长度时,这时的 s 子字符串就是 p 的异位词。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
61-
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class Solution:
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def findAnagrams(self, s: str, p: str) -> List[int]:
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counter = collections.Counter(p)
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res = []
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left = right = 0
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t = collections.Counter()
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while right < len(s):
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t[s[right]] += 1
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while t[s[right]] > counter[s[right]]:
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t[s[left]] -= 1
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left += 1
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if right - left == len(p) - 1:
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res.append(left)
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right += 1
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return res
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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"暴力"求解。利用哈希表 counter 统计字符串 p 中每个字符出现的次数。然后遍历字符串 s,判断子串 `s[i, i + p)` 中每个字符出现的次数是否与 counter 相同。若是,则将当前下标 i 添加到结果列表中。时间复杂度 `O(s.length * p.length)`。
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```java
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class Solution {
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public List<Integer> findAnagrams(String s, String p) {
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int[] counter = new int[26];
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for (int i = 0; i < p.length(); ++i) {
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++counter[p.charAt(i) - 'a'];
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}
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List<Integer> res = new ArrayList<>();
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for (int i = 0; i <= s.length() - p.length(); ++i) {
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int[] t = Arrays.copyOf(counter, counter.length);
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boolean find = true;
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for (int j = i; j < i + p.length(); ++j) {
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if (--t[s.charAt(j) - 'a'] < 0) {
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find = false;
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break;
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}
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}
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if (find) {
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res.add(i);
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}
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}
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return res;
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}
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}
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```
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"双指针 + 滑动窗口"求解。
114+
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```java
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class Solution {
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public List<Integer> findAnagrams(String s, String p) {
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int[] counter = new int[26];
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for (int i = 0; i < p.length(); ++i) {
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++counter[p.charAt(i) - 'a'];
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}
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List<Integer> res = new ArrayList<>();
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int left = 0, right = 0;
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int[] t = new int[26];
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while (right < s.length()) {
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int i = s.charAt(right) - 'a';
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++t[i];
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while (t[i] > counter[i]) {
129+
--t[s.charAt(left) - 'a'];
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++left;
131+
}
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if (right - left == p.length() - 1) {
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res.add(left);
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}
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++right;
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}
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return res;
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}
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}
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```
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### **...**

‎solution/0400-0499/0438.Find All Anagrams in a String/README_EN.md‎

Lines changed: 39 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -44,13 +44,50 @@ The substring with start index = 2 is &quot;ab&quot;, which is an anagram of &qu
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### **Python3**
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```python
47-
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class Solution:
48+
def findAnagrams(self, s: str, p: str) -> List[int]:
49+
counter = collections.Counter(p)
50+
res = []
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left = right = 0
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t = collections.Counter()
53+
while right < len(s):
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t[s[right]] += 1
55+
while t[s[right]] > counter[s[right]]:
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t[s[left]] -= 1
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left += 1
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if right - left == len(p) - 1:
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res.append(left)
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right += 1
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return res
4862
```
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### **Java**
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```java
53-
67+
class Solution {
68+
public List<Integer> findAnagrams(String s, String p) {
69+
int[] counter = new int[26];
70+
for (int i = 0; i < p.length(); ++i) {
71+
++counter[p.charAt(i) - 'a'];
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}
73+
List<Integer> res = new ArrayList<>();
74+
int left = 0, right = 0;
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int[] t = new int[26];
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while (right < s.length()) {
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int i = s.charAt(right) - 'a';
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++t[i];
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while (t[i] > counter[i]) {
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--t[s.charAt(left) - 'a'];
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++left;
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}
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if (right - left == p.length() - 1) {
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res.add(left);
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}
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++right;
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}
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return res;
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}
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}
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```
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### **...**
Lines changed: 24 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,24 @@
1+
class Solution {
2+
public List<Integer> findAnagrams(String s, String p) {
3+
int[] counter = new int[26];
4+
for (int i = 0; i < p.length(); ++i) {
5+
++counter[p.charAt(i) - 'a'];
6+
}
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List<Integer> res = new ArrayList<>();
8+
int left = 0, right = 0;
9+
int[] t = new int[26];
10+
while (right < s.length()) {
11+
int i = s.charAt(right) - 'a';
12+
++t[i];
13+
while (t[i] > counter[i]) {
14+
--t[s.charAt(left) - 'a'];
15+
++left;
16+
}
17+
if (right - left == p.length() - 1) {
18+
res.add(left);
19+
}
20+
++right;
21+
}
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return res;
23+
}
24+
}
Lines changed: 14 additions & 22 deletions
Original file line numberDiff line numberDiff line change
@@ -1,23 +1,15 @@
11
class Solution:
2-
def findAnagrams(self, s, p):
3-
"""
4-
:type s: str
5-
:type p: str
6-
:rtype: List[int]
7-
"""
8-
lens=len(s)
9-
lenp=len(p)
10-
if lens < lenp:
11-
return []
12-
flag1 = [0] * 26
13-
flag2 = [0] * 26
14-
for i, x in enumerate(p):
15-
flag1[ord(x) - 97] += 1
16-
ans = []
17-
for i, x in enumerate(s):
18-
flag2[ord(x)- 97] += 1
19-
if i >= lenp:
20-
flag2[ord(s[i - lenp]) - 97] -= 1
21-
if flag1 == flag2:
22-
ans.append(i-lenp+1)
23-
return ans
2+
def findAnagrams(self, s: str, p: str) -> List[int]:
3+
counter = collections.Counter(p)
4+
res = []
5+
left = right = 0
6+
t = collections.Counter()
7+
while right < len(s):
8+
t[s[right]] += 1
9+
while t[s[right]] > counter[s[right]]:
10+
t[s[left]] -= 1
11+
left += 1
12+
if right - left == len(p) - 1:
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res.append(left)
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right += 1
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return res

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