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Mar 13, 2025
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feat: add solutions to lc problem: No.1062 #4158

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114 changes: 78 additions & 36 deletions solution/1000-1099/1062.Longest Repeating Substring/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -66,20 +66,21 @@ tags:

### 方法一：动态规划

定义 $dp[i][j]$ 表示以 $s[i]$ 和 $s[j]$ 结尾的最长重复子串 🔒 的长度。状态转移方程为：
我们定义 $f[i][j]$ 表示以 $s[i]$ 和 $s[j]$ 结尾的最长重复子串的长度，初始时 $f[i][j]=0$。

我们在 $[1, n)$ 的区间内枚举 $i$，在 $[0, i)$ 的区间内枚举 $j$，如果 $s[i]=s[j]$，那么有：

$$
dp[i][j]=
f[i][j]=
\begin{cases}
dp[i-1][j-1]+1, & i>0 \cap s[i]=s[j] \\
1, & i=0 \cap s[i]=s[j] \\
0, & s[i] \neq s[j]
f[i-1][j-1]+1, & j>0 \\
1, & j=0
\end{cases}
$$

时间复杂度 $O(n^2)$，空间复杂度 $O(n^2)$。
我们求出所有 $f[i][j]$ 的最大值即为答案。

其中 $n$ 为字符串 $s$ 的长度。
时间复杂度 $O(n^2)$，空间复杂度 $O(n^2)$。其中 $n$ 为字符串 $s$ 的长度。

相似题目：

Expand All @@ -93,13 +94,13 @@ $$
class Solution:
def longestRepeatingSubstring(self, s: str) -> int:
n = len(s)
dp = [[0] * n for _ in range(n)]
f = [[0] * n for _ in range(n)]
ans = 0
for i in range(n):
for j in range(i + 1, n):
for i in range(1, n):
for j in range(i):
if s[i] == s[j]:
dp[i][j] = dp[i - 1][j - 1] + 1 if i else 1
ans = max(ans, dp[i][j])
f[i][j] = 1 + (f[i - 1][j - 1] if j else 0)
ans = max(ans, f[i][j])
return ans
```

Expand All @@ -109,13 +110,13 @@ class Solution:
class Solution {
public int longestRepeatingSubstring(String s) {
int n = s.length();
int[][] f = new int[n][n];
int ans = 0;
int[][] dp = new int[n][n];
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
for (int i = 1; i < n; ++i) {
for (int j = 0; j < i; ++j) {
if (s.charAt(i) == s.charAt(j)) {
dp[i][j] = i > 0 ? dp[i - 1][j - 1] + 1 : 1;
ans = Math.max(ans, dp[i][j]);
f[i][j] = 1 + (j > 0 ? f[i - 1][j - 1] : 0);
ans = Math.max(ans, f[i][j]);
}
}
}
Expand All @@ -130,14 +131,15 @@ class Solution {
class Solution {
public:
int longestRepeatingSubstring(string s) {
int n = s.size();
vector<vector<int>> dp(n, vector<int>(n));
int n = s.length();
int f[n][n];
memset(f, 0, sizeof(f));
int ans = 0;
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
for (int i = 1; i < n; ++i) {
for (int j = 0; j < i; ++j) {
if (s[i] == s[j]) {
dp[i][j] = i ? dp[i - 1][j - 1] + 1 : 1;
ans = max(ans, dp[i][j]);
f[i][j] = 1 + (j > 0 ? f[i - 1][j - 1] : 0);
ans = max(ans, f[i][j]);
}
}
}
Expand All @@ -149,26 +151,66 @@ public:
#### Go

```go
func longestRepeatingSubstring(s string) int {
func longestRepeatingSubstring(s string) (ans int) {
n := len(s)
dp := make([][]int, n)
for i := range dp {
dp[i] = make([]int, n)
f := make([][]int, n)
for i := range f {
f[i] = make([]int, n)
}
ans := 0
for i := 0; i < n; i++ {
for j := i + 1; j < n; j++ {
for i := 1; i < n; i++ {
for j := 0; j < i; j++ {
if s[i] == s[j] {
if i == 0 {
dp[i][j] = 1
} else {
dp[i][j] = dp[i-1][j-1] + 1
if j > 0 {
f[i][j] = f[i-1][j-1]
}
ans = max(ans, dp[i][j])
f[i][j]++
ans = max(ans, f[i][j])
}
}
}
return ans
return
}
```

#### TypeScript

```ts
function longestRepeatingSubstring(s: string): number {
const n = s.length;
const f: number[][] = Array.from({ length: n }).map(() => Array(n).fill(0));
let ans = 0;
for (let i = 1; i < n; ++i) {
for (let j = 0; j < i; ++j) {
if (s[i] === s[j]) {
f[i][j] = 1 + (f[i - 1][j - 1] || 0);
ans = Math.max(ans, f[i][j]);
}
}
}
return ans;
}
```

#### Rust

```rust
impl Solution {
pub fn longest_repeating_substring(s: String) -> i32 {
let n = s.len();
let mut f = vec![vec![0; n]; n];
let mut ans = 0;
let s = s.as_bytes();

for i in 1..n {
for j in 0..i {
if s[i] == s[j] {
f[i][j] = if j > 0 { f[i - 1][j - 1] + 1 } else { 1 };
ans = ans.max(f[i][j]);
}
}
}
ans
}
}
```

Expand Down
121 changes: 91 additions & 30 deletions solution/1000-1099/1062.Longest Repeating Substring/README_EN.md
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Expand Up @@ -62,7 +62,27 @@ tags:

<!-- solution:start -->

### Solution 1
### Solution 1: Dynamic Programming

We define $f[i][j]$ to represent the length of the longest repeating substring ending with $s[i]$ and $s[j]$. Initially, $f[i][j]=0$.

We enumerate $i$ in the range $[1, n)$ and enumerate $j$ in the range $[0, i)$. If $s[i]=s[j]$, then we have:

$$
f[i][j]=
\begin{cases}
f[i-1][j-1]+1, & j>0 \\
1, & j=0
\end{cases}
$$

The answer is the maximum value of all $f[i][j]$.

The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$. Where $n$ is the length of the string $s$.

Similar problems:

- [1044. Longest Duplicate Substring 🔒](https://github.com/doocs/leetcode/blob/main/solution/1000-1099/1044.Longest%20Duplicate%20Substring/README_EN.md)

<!-- tabs:start -->

Expand All @@ -72,13 +92,13 @@ tags:
class Solution:
def longestRepeatingSubstring(self, s: str) -> int:
n = len(s)
dp = [[0] * n for _ in range(n)]
f = [[0] * n for _ in range(n)]
ans = 0
for i in range(n):
for j in range(i + 1, n):
for i in range(1, n):
for j in range(i):
if s[i] == s[j]:
dp[i][j] = dp[i - 1][j - 1] + 1 if i else 1
ans = max(ans, dp[i][j])
f[i][j] = 1 + (f[i - 1][j - 1] if j else 0)
ans = max(ans, f[i][j])
return ans
```

Expand All @@ -88,13 +108,13 @@ class Solution:
class Solution {
public int longestRepeatingSubstring(String s) {
int n = s.length();
int[][] f = new int[n][n];
int ans = 0;
int[][] dp = new int[n][n];
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
for (int i = 1; i < n; ++i) {
for (int j = 0; j < i; ++j) {
if (s.charAt(i) == s.charAt(j)) {
dp[i][j] = i > 0 ? dp[i - 1][j - 1] + 1 : 1;
ans = Math.max(ans, dp[i][j]);
f[i][j] = 1 + (j > 0 ? f[i - 1][j - 1] : 0);
ans = Math.max(ans, f[i][j]);
}
}
}
Expand All @@ -109,14 +129,15 @@ class Solution {
class Solution {
public:
int longestRepeatingSubstring(string s) {
int n = s.size();
vector<vector<int>> dp(n, vector<int>(n));
int n = s.length();
int f[n][n];
memset(f, 0, sizeof(f));
int ans = 0;
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
for (int i = 1; i < n; ++i) {
for (int j = 0; j < i; ++j) {
if (s[i] == s[j]) {
dp[i][j] = i ? dp[i - 1][j - 1] + 1 : 1;
ans = max(ans, dp[i][j]);
f[i][j] = 1 + (j > 0 ? f[i - 1][j - 1] : 0);
ans = max(ans, f[i][j]);
}
}
}
Expand All @@ -128,26 +149,66 @@ public:
#### Go

```go
func longestRepeatingSubstring(s string) int {
func longestRepeatingSubstring(s string) (ans int) {
n := len(s)
dp := make([][]int, n)
for i := range dp {
dp[i] = make([]int, n)
f := make([][]int, n)
for i := range f {
f[i] = make([]int, n)
}
ans := 0
for i := 0; i < n; i++ {
for j := i + 1; j < n; j++ {
for i := 1; i < n; i++ {
for j := 0; j < i; j++ {
if s[i] == s[j] {
if i == 0 {
dp[i][j] = 1
} else {
dp[i][j] = dp[i-1][j-1] + 1
if j > 0 {
f[i][j] = f[i-1][j-1]
}
ans = max(ans, dp[i][j])
f[i][j]++
ans = max(ans, f[i][j])
}
}
}
return ans
return
}
```

#### TypeScript

```ts
function longestRepeatingSubstring(s: string): number {
const n = s.length;
const f: number[][] = Array.from({ length: n }).map(() => Array(n).fill(0));
let ans = 0;
for (let i = 1; i < n; ++i) {
for (let j = 0; j < i; ++j) {
if (s[i] === s[j]) {
f[i][j] = 1 + (f[i - 1][j - 1] || 0);
ans = Math.max(ans, f[i][j]);
}
}
}
return ans;
}
```

#### Rust

```rust
impl Solution {
pub fn longest_repeating_substring(s: String) -> i32 {
let n = s.len();
let mut f = vec![vec![0; n]; n];
let mut ans = 0;
let s = s.as_bytes();

for i in 1..n {
for j in 0..i {
if s[i] == s[j] {
f[i][j] = if j > 0 { f[i - 1][j - 1] + 1 } else { 1 };
ans = ans.max(f[i][j]);
}
}
}
ans
}
}
```

Expand Down
15 changes: 8 additions & 7 deletions solution/1000-1099/1062.Longest Repeating Substring/Solution.cpp
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Original file line number Diff line number Diff line change
@@ -1,17 +1,18 @@
class Solution {
public:
int longestRepeatingSubstring(string s) {
int n = s.size();
vector<vector<int>> dp(n, vector<int>(n));
int n = s.length();
int f[n][n];
memset(f, 0, sizeof(f));
int ans = 0;
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
for (int i = 1; i < n; ++i) {
for (int j = 0; j < i; ++j) {
if (s[i] == s[j]) {
dp[i][j] = i ? dp[i - 1][j - 1] + 1 : 1;
ans = max(ans, dp[i][j]);
f[i][j] = 1 + (j > 0 ? f[i - 1][j - 1] : 0);
ans = max(ans, f[i][j]);
}
}
}
return ans;
}
};
};
26 changes: 12 additions & 14 deletions solution/1000-1099/1062.Longest Repeating Substring/Solution.go
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Original file line number Diff line number Diff line change
@@ -1,21 +1,19 @@
func longestRepeatingSubstring(s string) int {
func longestRepeatingSubstring(s string) (ans int) {
n := len(s)
dp := make([][]int, n)
for i := range dp {
dp[i] = make([]int, n)
f := make([][]int, n)
for i := range f {
f[i] = make([]int, n)
}
ans := 0
for i := 0; i < n; i++ {
for j := i + 1; j < n; j++ {
for i := 1; i < n; i++ {
for j := 0; j < i; j++ {
if s[i] == s[j] {
if i == 0 {
dp[i][j] = 1
} else {
dp[i][j] = dp[i-1][j-1] + 1
if j > 0 {
f[i][j] = f[i-1][j-1]
}
ans = max(ans, dp[i][j])
f[i][j]++
ans = max(ans, f[i][j])
}
}
}
return ans
}
return
}
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