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Commit 138ef72

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feat: update solutions to lc problem: No.0187
No.0187.Repeated DNA Sequences
1 parent 7f17820 commit 138ef72

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3 files changed

+86
-19
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3 files changed

+86
-19
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‎solution/0100-0199/0187.Repeated DNA Sequences/README.md‎

Lines changed: 37 additions & 6 deletions
Original file line numberDiff line numberDiff line change
@@ -45,9 +45,17 @@
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<!-- 这里可写通用的实现逻辑 -->
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**朴素解法:**
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**方法一:哈希表**
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使用哈希表,记录所有连续长度为 10 的子字符串出现次数(字符串为 Key,次数为 Value),当出现一次以上时,将其加入返回列表当中。
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朴素解法,用哈希表保存所有长度为 10 的子序列出现的次数,当子序列出现次数大于 1 时,把该子序列作为结果之一。
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假设字符串 `s` 长度为 `n`,则时间复杂度 $O(n \times 10),ドル空间复杂度 $O(n)$。
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**方法二:Rabin-Karp 字符串匹配算法**
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本质上是滑动窗口和哈希的结合方法,和 [0028.找出字符串中第一个匹配项的下标](https://leetcode.cn/problems/find-the-index-of-the-first-occurrence-in-a-string/) 类似,本题可以借助哈希函数将子序列计数的时间复杂度降低到 $O(1)$。
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假设字符串 `s` 长度为 `n`,则时间复杂度为 $O(n),ドル空间复杂度 $O(n)$。
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<!-- tabs:start -->
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@@ -115,12 +123,12 @@ var findRepeatedDnaSequences = function (s) {
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### **Go**
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哈希表:
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```go
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func findRepeatedDnaSequences(s string) []string {
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cnt := make(map[string]int)
121-
n := len(s) - 10
122-
ans := make([]string, 0)
123-
for i := 0; i <= n; i++ {
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ans, cnt := []string{}, map[string]int{}
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for i := 0; i <= len(s)-10; i++ {
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sub := s[i : i+10]
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cnt[sub]++
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if cnt[sub] == 2 {
@@ -131,6 +139,29 @@ func findRepeatedDnaSequences(s string) []string {
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}
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```
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Rabin-Karp:
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```go
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func findRepeatedDnaSequences(s string) []string {
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hashCode := map[byte]int{'A': 0, 'C': 1, 'G': 2, 'T': 3}
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ans, cnt, left, right := []string{}, map[int]int{}, 0, 0
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sha, multi := 0, int(math.Pow(4, 9))
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for ; right < len(s); right++ {
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sha = sha*4 + hashCode[s[right]]
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if right-left+1 < 10 {
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continue
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}
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cnt[sha]++
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if cnt[sha] == 2 {
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ans = append(ans, s[left:right+1])
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}
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sha, left = sha-multi*hashCode[s[left]], left+1
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}
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return ans
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}
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```
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### **C++**
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```cpp

‎solution/0100-0199/0187.Repeated DNA Sequences/README_EN.md‎

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Original file line numberDiff line numberDiff line change
@@ -32,6 +32,14 @@
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## Solutions
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**Approach 1: HashTable**
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Time complexity $O(n \times 10),ドル Space complexity $O(n)$.
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**Approach 2: Rabin-Karp**
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Time complexity $O(n),ドル Space complexity $O(n)$.
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<!-- tabs:start -->
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### **Python3**
@@ -94,12 +102,12 @@ var findRepeatedDnaSequences = function (s) {
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### **Go**
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HashTable:
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```go
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func findRepeatedDnaSequences(s string) []string {
99-
cnt := make(map[string]int)
100-
n := len(s) - 10
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ans := make([]string, 0)
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for i := 0; i <= n; i++ {
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ans, cnt := []string{}, map[string]int{}
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for i := 0; i <= len(s)-10; i++ {
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sub := s[i : i+10]
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cnt[sub]++
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if cnt[sub] == 2 {
@@ -110,6 +118,29 @@ func findRepeatedDnaSequences(s string) []string {
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}
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```
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Rabin-Karp:
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```go
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func findRepeatedDnaSequences(s string) []string {
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hashCode := map[byte]int{'A': 0, 'C': 1, 'G': 2, 'T': 3}
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ans, cnt, left, right := []string{}, map[int]int{}, 0, 0
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sha, multi := 0, int(math.Pow(4, 9))
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for ; right < len(s); right++ {
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sha = sha*4 + hashCode[s[right]]
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if right-left+1 < 10 {
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continue
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}
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cnt[sha]++
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if cnt[sha] == 2 {
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ans = append(ans, s[left:right+1])
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}
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sha, left = sha-multi*hashCode[s[left]], left+1
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}
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return ans
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}
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```
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### **C++**
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```cpp
Lines changed: 14 additions & 9 deletions
Original file line numberDiff line numberDiff line change
@@ -1,13 +1,18 @@
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func findRepeatedDnaSequences(s string) []string {
2-
cnt := make(map[string]int)
3-
n:= len(s) -10
4-
ans:=make([]string, 0)
5-
fori := 0; i<=n; i++ {
6-
sub:=s[i : i+10]
7-
cnt[sub]++
8-
if cnt[sub] ==2 {
9-
ans=append(ans, sub)
2+
hashCode := map[byte]int{'A': 0, 'C': 1, 'G': 2, 'T': 3}
3+
ans, cnt, left, right:= []string{}, map[int]int{}, 0, 0
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5+
sha, multi := 0, int(math.Pow(4, 9))
6+
for ; right<len(s); right++ {
7+
sha=sha*4+hashCode[s[right]]
8+
if right-left+1<10 {
9+
continue
1010
}
11+
cnt[sha]++
12+
if cnt[sha] == 2 {
13+
ans = append(ans, s[left:right+1])
14+
}
15+
sha, left = sha-multi*hashCode[s[left]], left+1
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}
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return ans
13-
}
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}

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