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feat: add solutions to lcci problems: No.08.06~08.09 #4068

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Feb 15, 2025
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feat: add solutions to lcci problems: No.08.06~08.09 #4068

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2 changes: 1 addition & 1 deletion lcci/08.06.Hanota/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -91,7 +91,7 @@ class Solution {
class Solution {
public:
void hanota(vector<int>& A, vector<int>& B, vector<int>& C) {
function<void(int, vector<int>&, vector<int>&, vector<int>&)> dfs = [&](int n, vector<int>& a, vector<int>& b, vector<int>& c) {
auto dfs = [&](this auto&& dfs, int n, vector<int>& a, vector<int>& b, vector<int>& c) {
if (n == 1) {
c.push_back(a.back());
a.pop_back();
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2 changes: 1 addition & 1 deletion lcci/08.06.Hanota/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -98,7 +98,7 @@ class Solution {
class Solution {
public:
void hanota(vector<int>& A, vector<int>& B, vector<int>& C) {
function<void(int, vector<int>&, vector<int>&, vector<int>&)> dfs = [&](int n, vector<int>& a, vector<int>& b, vector<int>& c) {
auto dfs = [&](this auto&& dfs, int n, vector<int>& a, vector<int>& b, vector<int>& c) {
if (n == 1) {
c.push_back(a.back());
a.pop_back();
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4 changes: 2 additions & 2 deletions lcci/08.06.Hanota/Solution.cpp
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Original file line number Diff line number Diff line change
@@ -1,7 +1,7 @@
class Solution {
public:
void hanota(vector<int>& A, vector<int>& B, vector<int>& C) {
function<void(int, vector<int>&, vector<int>&, vector<int>&)> dfs = [&](int n, vector<int>& a, vector<int>& b, vector<int>& c) {
auto dfs = [&](this auto&& dfs, int n, vector<int>& a, vector<int>& b, vector<int>& c) {
if (n == 1) {
c.push_back(a.back());
a.pop_back();
Expand All @@ -14,4 +14,4 @@ class Solution {
};
dfs(A.size(), A, B, C);
}
};
};
138 changes: 65 additions & 73 deletions lcci/08.07.Permutation I/README.md
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Expand Up @@ -45,7 +45,7 @@ edit_url: https://github.com/doocs/leetcode/edit/main/lcci/08.07.Permutation%20I

### 方法一：DFS（回溯）

我们设计一个函数 $dfs(i)$ 表示已经填完了前 $i$ 个位置，现在需要填第 $i+1$ 个位置。枚举所有可能的字符，如果这个字符没有被填过，就填入这个字符，然后继续填下一个位置，直到填完所有的位置。
我们设计一个函数 $\textit{dfs}(i)$ 表示已经填完了前 $i$ 个位置，现在需要填第 $i+1$ 个位置。枚举所有可能的字符，如果这个字符没有被填过，就填入这个字符，然后继续填下一个位置，直到填完所有的位置。

时间复杂度 $O(n \times n!)$，其中 $n$ 是字符串的长度。一共有 $n!$ 个排列，每个排列需要 $O(n)$ 的时间来构造。

Expand All @@ -57,22 +57,20 @@ edit_url: https://github.com/doocs/leetcode/edit/main/lcci/08.07.Permutation%20I
class Solution:
def permutation(self, S: str) -> List[str]:
def dfs(i: int):
if i == n:
if i >= n:
ans.append("".join(t))
return
for j, c in enumerate(S):
if vis[j]:
continue
vis[j] = True
t.append(c)
dfs(i + 1)
t.pop()
vis[j] = False
if not vis[j]:
vis[j] = True
t[i] = c
dfs(i + 1)
vis[j] = False

ans = []
n = len(S)
vis = [False] * n
ans = []
t = []
t = list(S)
dfs(0)
return ans
```
Expand All @@ -82,30 +80,31 @@ class Solution:
```java
class Solution {
private char[] s;
private boolean[] vis = new boolean['z' + 1];
private char[] t;
private boolean[] vis;
private List<String> ans = new ArrayList<>();
private StringBuilder t = new StringBuilder();

public String[] permutation(String S) {
s = S.toCharArray();
int n = s.length;
vis = new boolean[n];
t = new char[n];
dfs(0);
return ans.toArray(new String[0]);
}

private void dfs(int i) {
if (i == s.length) {
ans.add(t.toString());
if (i >= s.length) {
ans.add(new String(t));
return;
}
for (char c : s) {
if (vis[c]) {
continue;
for (int j = 0; j < s.length; ++j) {
if (!vis[j]) {
vis[j] = true;
t[i] = s[j];
dfs(i + 1);
vis[j] = false;
}
vis[c] = true;
t.append(c);
dfs(i + 1);
t.deleteCharAt(t.length() - 1);
vis[c] = false;
}
}
}
Expand All @@ -119,51 +118,49 @@ public:
vector<string> permutation(string S) {
int n = S.size();
vector<bool> vis(n);
string t = S;
vector<string> ans;
string t;
function<void(int)> dfs = [&](int i) {
auto dfs = [&](this auto&& dfs, int i) {
if (i >= n) {
ans.push_back(t);
ans.emplace_back(t);
return;
}
for (int j = 0; j < n; ++j) {
if (vis[j]) {
continue;
if (!vis[j]) {
vis[j] = true;
t[i] = S[j];
dfs(i + 1);
vis[j] = false;
}
vis[j] = true;
t.push_back(S[j]);
dfs(i + 1);
t.pop_back();
vis[j] = false;
}
};
dfs(0);
return ans;
}
};

```

#### Go

```go
func permutation(S string) (ans []string) {
t := []byte{}
vis := make([]bool, len(S))
t := []byte(S)
n := len(t)
vis := make([]bool, n)
var dfs func(int)
dfs = func(i int) {
if i >= len(S) {
if i >= n {
ans = append(ans, string(t))
return
}
for j := range S {
if vis[j] {
continue
if !vis[j] {
vis[j] = true
t[i] = S[j]
dfs(i + 1)
vis[j] = false
}
vis[j] = true
t = append(t, S[j])
dfs(i + 1)
t = t[:len(t)-1]
vis[j] = false
}
}
dfs(0)
Expand All @@ -178,7 +175,7 @@ function permutation(S: string): string[] {
const n = S.length;
const vis: boolean[] = Array(n).fill(false);
const ans: string[] = [];
const t: string[] = [];
const t: string[] = Array(n).fill('');
const dfs = (i: number) => {
if (i >= n) {
ans.push(t.join(''));
Expand All @@ -189,9 +186,8 @@ function permutation(S: string): string[] {
continue;
}
vis[j] = true;
t.push(S[j]);
t[i] = S[j];
dfs(i + 1);
t.pop();
vis[j] = false;
}
};
Expand All @@ -211,7 +207,7 @@ var permutation = function (S) {
const n = S.length;
const vis = Array(n).fill(false);
const ans = [];
const t = [];
const t = Array(n).fill('');
const dfs = i => {
if (i >= n) {
ans.push(t.join(''));
Expand All @@ -222,9 +218,8 @@ var permutation = function (S) {
continue;
}
vis[j] = true;
t.push(S[j]);
t[i] = S[j];
dfs(i + 1);
t.pop();
vis[j] = false;
}
};
Expand All @@ -237,33 +232,30 @@ var permutation = function (S) {

```swift
class Solution {
private var s: [Character] = []
private var vis: [Bool] = Array(repeating: false, count: 128)
private var ans: [String] = []
private var t: String = ""

func permutation(_ S: String) -> [String] {
s = Array(S)
dfs(0)
return ans
}

private func dfs(_ i: Int) {
if i == s.count {
ans.append(t)
return
}
for c in s {
let index = Int(c.asciiValue!)
if vis[index] {
continue
var ans: [String] = []
let s = Array(S)
var t = s
var vis = Array(repeating: false, count: s.count)
let n = s.count

func dfs(_ i: Int) {
if i >= n {
ans.append(String(t))
return
}
for j in 0..<n {
if !vis[j] {
vis[j] = true
t[i] = s[j]
dfs(i + 1)
vis[j] = false
}
}
vis[index] = true
t.append(c)
dfs(i + 1)
t.removeLast()
vis[index] = false
}

dfs(0)
return ans
}
}
```
Expand Down
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