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feat: add solutions to lc problem: No.0830
No.0830.Positions of Large Groups
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‎solution/0800-0899/0830.Positions of Large Groups/README.md

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@@ -59,22 +59,100 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：双指针**
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我们用双指针 $i$ 和 $j$ 找到每个分组的起始位置和终止位置，然后判断分组长度是否大于等于 $3$，若是则将其加入结果数组。
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时间复杂度 $O(n)$，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 largeGroupPositions(self, s: str) -> List[List[int]]:
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i, n = 0, len(s)
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ans = []
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while i < n:
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j = i
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while j < n and s[j] == s[i]:
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j += 1
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if j - i >= 3:
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ans.append([i, j - 1])
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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 List<List<Integer>> largeGroupPositions(String s) {
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int n = s.length();
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int i = 0;
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List<List<Integer>> ans = new ArrayList<>();
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while (i < n) {
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int j = i;
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while (j < n && s.charAt(j) == s.charAt(i)) {
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++j;
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}
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if (j - i >= 3) {
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ans.add(Arrays.asList(i, j - 1));
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}
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i = j;
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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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vector<vector<int>> largeGroupPositions(string s) {
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int n = s.size();
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int i = 0;
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vector<vector<int>> ans;
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while (i < n) {
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int j = i;
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while (j < n && s[j] == s[i]) {
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++j;
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}
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if (j - i >= 3) {
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ans.push_back({i, j - 1});
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}
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i = j;
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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 largeGroupPositions(s string) [][]int {
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i, n := 0, len(s)
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ans := [][]int{}
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for i < n {
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j := i
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for j < n && s[j] == s[i] {
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j++
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}
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if j-i >= 3 {
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ans = append(ans, []int{i, j - 1})
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}
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i = j
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}
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return ans
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}
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```
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### **...**

‎solution/0800-0899/0830.Positions of Large Groups/README_EN.md

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@@ -54,13 +54,85 @@
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### **Python3**
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```python
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class Solution:
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def largeGroupPositions(self, s: str) -> List[List[int]]:
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i, n = 0, len(s)
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ans = []
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while i < n:
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j = i
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while j < n and s[j] == s[i]:
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j += 1
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if j - i >= 3:
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ans.append([i, j - 1])
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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 List<List<Integer>> largeGroupPositions(String s) {
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int n = s.length();
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int i = 0;
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List<List<Integer>> ans = new ArrayList<>();
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while (i < n) {
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int j = i;
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while (j < n && s.charAt(j) == s.charAt(i)) {
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++j;
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}
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if (j - i >= 3) {
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ans.add(Arrays.asList(i, j - 1));
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}
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i = j;
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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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vector<vector<int>> largeGroupPositions(string s) {
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int n = s.size();
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int i = 0;
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vector<vector<int>> ans;
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while (i < n) {
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int j = i;
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while (j < n && s[j] == s[i]) {
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++j;
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}
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if (j - i >= 3) {
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ans.push_back({i, j - 1});
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}
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i = j;
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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 largeGroupPositions(s string) [][]int {
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i, n := 0, len(s)
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ans := [][]int{}
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for i < n {
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j := i
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for j < n && s[j] == s[i] {
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j++
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}
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if j-i >= 3 {
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ans = append(ans, []int{i, j - 1})
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}
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i = j
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}
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return ans
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}
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```
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### **...**

‎solution/0800-0899/0830.Positions of Large Groups/Solution.cpp

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class Solution {
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public:
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vector<vector<int>> largeGroupPositions(string s) {
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int n = s.size();
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int i = 0;
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vector<vector<int>> ans;
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while (i < n) {
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int j = i;
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while (j < n && s[j] == s[i]) {
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++j;
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}
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if (j - i >= 3) {
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ans.push_back({i, j - 1});
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}
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i = j;
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}
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return ans;
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}
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};

‎solution/0800-0899/0830.Positions of Large Groups/Solution.go

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func largeGroupPositions(s string) [][]int {
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i, n := 0, len(s)
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ans := [][]int{}
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for i < n {
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j := i
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for j < n && s[j] == s[i] {
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j++
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}
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if j-i >= 3 {
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ans = append(ans, []int{i, j - 1})
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}
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i = j
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}
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return ans
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}

‎solution/0800-0899/0830.Positions of Large Groups/Solution.java

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class Solution {
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public List<List<Integer>> largeGroupPositions(String s) {
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int n = s.length();
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int i = 0;
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List<List<Integer>> ans = new ArrayList<>();
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while (i < n) {
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int j = i;
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while (j < n && s.charAt(j) == s.charAt(i)) {
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++j;
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}
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if (j - i >= 3) {
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ans.add(Arrays.asList(i, j - 1));
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}
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i = j;
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}
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return ans;
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}
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}

‎solution/0800-0899/0830.Positions of Large Groups/Solution.py

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class Solution:
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def largeGroupPositions(self, s: str) -> List[List[int]]:
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i, n = 0, len(s)
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ans = []
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while i < n:
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j = i
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while j < n and s[j] == s[i]:
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j += 1
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if j - i >= 3:
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ans.append([i, j - 1])
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i = j
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return ans

‎solution/0900-0999/0901.Online Stock Span/README.md

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@@ -53,7 +53,7 @@ S.next(85) 被调用并返回 6。
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这实际上是经典的单调栈模型，找出左侧第一个比当前元素大的元素。
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我们维护一个从栈底到栈顶单调递减的栈，栈中每个元素存放的是 `(price, cnt)` 数据对，其中 `price` 表示价格，`cnt` 表示当前价格的跨度。
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我们维护一个从栈底到栈顶价格单调递减的栈，栈中每个元素存放的是 `(price, cnt)` 数据对，其中 `price` 表示价格，`cnt` 表示当前价格的跨度。
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出现价格 `price` 时，我们将其与栈顶元素进行比较，如果栈顶元素的价格小于等于 `price`，则将当日价格的跨度 `cnt` 加上栈顶元素的跨度，然后将栈顶元素出栈，直到栈顶元素的价格大于 `price`，或者栈为空为止。
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StockSpanner() {
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}
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int next(int price) {
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int ans = 1;
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int cnt = 1;
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while (!stk.empty() && stk.top().first <= price) {
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ans += stk.top().second;
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cnt += stk.top().second;
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stk.pop();
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}
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stk.push({price, ans});
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return ans;
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stk.push({price, cnt});
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return cnt;
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}
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private:

‎solution/0900-0999/0901.Online Stock Span/README_EN.md

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@@ -107,7 +107,7 @@ public:
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StockSpanner() {
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}
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int next(int price) {
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int cnt = 1;
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while (!stk.empty() && stk.top().first <= price) {

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