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Commit 1240ef1

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feat: add solutions to lc problem: No.1062
No.1062.Longest Repeating Substring
1 parent 6fe9df5 commit 1240ef1

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7 files changed

+253
-4
lines changed

7 files changed

+253
-4
lines changed

‎solution/0800-0899/0801.Minimum Swaps To Make Sequences Increasing/README.md‎

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@@ -57,9 +57,9 @@ A = [1, 3, 5, 7] , B = [1, 2, 3, 4]
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**方法一:动态规划**
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定义 $a,ドル $b$ 分别表示使得前 $i$ 个元素序列严格递增,且第 $i$ 个元素不交换、交换的最小交换次数。
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定义 $a,ドル $b$ 分别表示使得下标 $[0..i]$ 的元素序列严格递增,且第 $i$ 个元素不交换、交换的最小交换次数。下标从 0ドル$ 开始。
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当 $i=0$ 时,易知 $a = 0,ドル $b=1$。
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当 $i=0$ 时,有 $a = 0,ドル $b=1$。
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当 $i\gt 0$ 时,我们先将此前 $a,ドル $b$ 的值保存在 $x,ドル $y$ 中,然后分情况讨论:
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‎solution/1000-1099/1062.Longest Repeating Substring/README.md‎

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Original file line numberDiff line numberDiff line change
@@ -50,22 +50,119 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:动态规划**
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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:
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$$
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dp[i][j]=
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\begin{cases}
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dp[i-1][j-1]+1, & i>0 \cap s[i]=s[j] \\
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1, & i=0 \cap s[i]=s[j] \\
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0, & s[i] \neq s[j]
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\end{cases}
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$$
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时间复杂度 $O(n^2),ドル空间复杂度 $O(n^2)$。
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其中 $n$ 为字符串 $s$ 的长度。
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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
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class Solution:
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def longestRepeatingSubstring(self, s: str) -> int:
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n = len(s)
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dp = [[0] * n for _ in range(n)]
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ans = 0
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for i in range(n):
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for j in range(i + 1, n):
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if s[i] == s[j]:
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dp[i][j] = dp[i - 1][j - 1] + 1 if i else 1
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ans = max(ans, dp[i][j])
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return ans
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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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```java
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class Solution {
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public int longestRepeatingSubstring(String s) {
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int n = s.length();
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int ans = 0;
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int[][] dp = new int[n][n];
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for (int i = 0; i < n; ++i) {
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for (int j = i + 1; j < n; ++j) {
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if (s.charAt(i) == s.charAt(j)) {
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dp[i][j] = i > 0 ? dp[i - 1][j - 1] + 1 : 1;
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ans = Math.max(ans, dp[i][j]);
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}
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}
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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 longestRepeatingSubstring(string s) {
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int n = s.size();
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vector<vector<int>> dp(n, vector<int>(n));
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int ans = 0;
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for (int i = 0; i < n; ++i) {
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for (int j = i + 1; j < n; ++j) {
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if (s[i] == s[j]) {
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dp[i][j] = i ? dp[i - 1][j - 1] + 1 : 1;
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ans = max(ans, dp[i][j]);
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}
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}
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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 longestRepeatingSubstring(s string) int {
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n := len(s)
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dp := make([][]int, n)
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for i := range dp {
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dp[i] = make([]int, n)
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}
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ans := 0
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for i := 0; i < n; i++ {
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for j := i + 1; j < n; j++ {
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if s[i] == s[j] {
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if i == 0 {
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dp[i][j] = 1
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} else {
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dp[i][j] = dp[i-1][j-1] + 1
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}
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ans = max(ans, dp[i][j])
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}
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}
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}
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return ans
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**

‎solution/1000-1099/1062.Longest Repeating Substring/README_EN.md‎

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Original file line numberDiff line numberDiff line change
@@ -46,13 +46,93 @@
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### **Python3**
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```python
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class Solution:
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def longestRepeatingSubstring(self, s: str) -> int:
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n = len(s)
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dp = [[0] * n for _ in range(n)]
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ans = 0
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for i in range(n):
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for j in range(i + 1, n):
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if s[i] == s[j]:
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dp[i][j] = dp[i - 1][j - 1] + 1 if i else 1
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ans = max(ans, dp[i][j])
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return ans
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```
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### **Java**
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```java
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class Solution {
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public int longestRepeatingSubstring(String s) {
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int n = s.length();
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int ans = 0;
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int[][] dp = new int[n][n];
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for (int i = 0; i < n; ++i) {
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for (int j = i + 1; j < n; ++j) {
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if (s.charAt(i) == s.charAt(j)) {
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dp[i][j] = i > 0 ? dp[i - 1][j - 1] + 1 : 1;
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ans = Math.max(ans, dp[i][j]);
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}
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}
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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:
88+
int longestRepeatingSubstring(string s) {
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int n = s.size();
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vector<vector<int>> dp(n, vector<int>(n));
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int ans = 0;
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for (int i = 0; i < n; ++i) {
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for (int j = i + 1; j < n; ++j) {
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if (s[i] == s[j]) {
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dp[i][j] = i ? dp[i - 1][j - 1] + 1 : 1;
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ans = max(ans, dp[i][j]);
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}
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}
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}
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return ans;
101+
}
102+
};
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```
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### **Go**
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```go
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func longestRepeatingSubstring(s string) int {
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n := len(s)
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dp := make([][]int, n)
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for i := range dp {
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dp[i] = make([]int, n)
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}
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ans := 0
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for i := 0; i < n; i++ {
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for j := i + 1; j < n; j++ {
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if s[i] == s[j] {
118+
if i == 0 {
119+
dp[i][j] = 1
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} else {
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dp[i][j] = dp[i-1][j-1] + 1
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}
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ans = max(ans, dp[i][j])
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}
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}
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}
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return ans
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**
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Original file line numberDiff line numberDiff line change
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class Solution {
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public:
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int longestRepeatingSubstring(string s) {
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int n = s.size();
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vector<vector<int>> dp(n, vector<int>(n));
6+
int ans = 0;
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for (int i = 0; i < n; ++i) {
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for (int j = i + 1; j < n; ++j) {
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if (s[i] == s[j]) {
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dp[i][j] = i ? dp[i - 1][j - 1] + 1 : 1;
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ans = max(ans, dp[i][j]);
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}
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}
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}
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return ans;
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}
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};
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Original file line numberDiff line numberDiff line change
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func longestRepeatingSubstring(s string) int {
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n := len(s)
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dp := make([][]int, n)
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for i := range dp {
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dp[i] = make([]int, n)
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}
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ans := 0
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for i := 0; i < n; i++ {
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for j := i + 1; j < n; j++ {
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if s[i] == s[j] {
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if i == 0 {
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dp[i][j] = 1
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} else {
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dp[i][j] = dp[i-1][j-1] + 1
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}
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ans = max(ans, dp[i][j])
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}
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}
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}
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return ans
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,16 @@
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class Solution {
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public int longestRepeatingSubstring(String s) {
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int n = s.length();
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int ans = 0;
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int[][] dp = new int[n][n];
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for (int i = 0; i < n; ++i) {
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for (int j = i + 1; j < n; ++j) {
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if (s.charAt(i) == s.charAt(j)) {
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dp[i][j] = i > 0 ? dp[i - 1][j - 1] + 1 : 1;
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ans = Math.max(ans, dp[i][j]);
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}
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}
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}
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return ans;
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}
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}
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,11 @@
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class Solution:
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def longestRepeatingSubstring(self, s: str) -> int:
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n = len(s)
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dp = [[0] * n for _ in range(n)]
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ans = 0
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for i in range(n):
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for j in range(i + 1, n):
8+
if s[i] == s[j]:
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dp[i][j] = dp[i - 1][j - 1] + 1 if i else 1
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ans = max(ans, dp[i][j])
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return ans

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