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feat: add solutions to lc problems
* No.2190.Most Frequent Number Following Key In an Array * No.2191.Sort the Jumbled Numbers * NO.2193.Minimum Number of Moves to Make Palindrome
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‎README.md

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### 6. 图论
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- [网络延迟时间](/solution/0700-0799/0743.Network%20Delay%20Time/README.md) - `最短路`、`Dijkstra 算法`、`Bellman Ford 算法`、`SPFA 算法`
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- [得到要求路径的最小带权子图](/solution/2200-2299/2203.Minimum%20Weighted%20Subgraph%20With%20the%20Required%20Paths/README.md) - `最短路`、`Dijkstra 算法`
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- [连接所有点的最小费用](/solution/1500-1599/1584.Min%20Cost%20to%20Connect%20All%20Points/README.md) - `最小生成树`、`Prim 算法`、`Kruskal 算法`
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- [最低成本联通所有城市](/solution/1100-1199/1135.Connecting%20Cities%20With%20Minimum%20Cost/README.md) - `最小生成树`、`Kruskal 算法`、`并查集`
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- [水资源分配优化](/solution/1100-1199/1168.Optimize%20Water%20Distribution%20in%20a%20Village/README.md) - `最小生成树`、`Kruskal 算法`、`并查集`

‎README_EN.md

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@@ -141,6 +141,7 @@ Complete solutions to [LeetCode](https://leetcode.com/problemset/all/), [LCOF](h
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### 6. Graph Theory
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- [Network Delay Time](/solution/0700-0799/0743.Network%20Delay%20Time/README_EN.md) - `Shortest Path`, `Dijkstra's algorithm`, `Bellman Ford's algorithm`, `SPFA`
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- [Minimum Weighted Subgraph With the Required Paths](/solution/2200-2299/2203.Minimum%20Weighted%20Subgraph%20With%20the%20Required%20Paths/README_EN.md) - `Shortest Path`, `Dijkstra's algorithm`
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- [Min Cost to Connect All Points](/solution/1500-1599/1584.Min%20Cost%20to%20Connect%20All%20Points/README_EN.md) - `Minimum Spanning Tree`, `Prim's algorithm`, `Kruskal's algorithm`, `Union find`
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- [Connecting Cities With Minimum Cost](/solution/1100-1199/1135.Connecting%20Cities%20With%20Minimum%20Cost/README_EN.md) - `Minimum Spanning Tree`, `Kruskal's algorithm`
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- [Optimize Water Distribution in a Village](/solution/1100-1199/1168.Optimize%20Water%20Distribution%20in%20a%20Village/README_EN.md) - `Minimum Spanning Tree`, `Kruskal's algorithm`, `Union find`

‎lcci/17.10.Find Majority Element/Solution.py

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@@ -5,5 +5,5 @@ def majorityElement(self, nums: List[int]) -> int:
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if cnt == 0:
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m, cnt = v, 1
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else:
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cnt += (1 if m == v else -1)
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cnt += 1 if m == v else -1
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return m if nums.count(m) > len(nums) // 2 else -1

‎lcof/面试题39. 数组中出现次数超过一半的数字/Solution.py

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@@ -5,5 +5,5 @@ def majorityElement(self, nums: List[int]) -> int:
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if cnt == 0:
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m, cnt = v, 1
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else:
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cnt += (1 if m == v else -1)
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cnt += 1 if m == v else -1
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return m

‎solution/0100-0199/0160.Intersection of Two Linked Lists/Solution.py

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# self.val = x
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# self.next = None
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class Solution:
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def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
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a, b = headA, headB

‎solution/0100-0199/0169.Majority Element/Solution.py

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if cnt == 0:
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m, cnt = v, 1
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else:
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cnt += (1 if m == v else -1)
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cnt += 1 if m == v else -1
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return m

‎solution/2100-2199/2190.Most Frequent Number Following Key In an Array/README.md

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@@ -58,15 +58,88 @@ target = 2 是紧跟着 key 之后出现次数最多的数字，所以我们返
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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 mostFrequent(self, nums: List[int], key: int) -> int:
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cnt = Counter()
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mx = ans = 0
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for i, v in enumerate(nums[:-1]):
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if v == key:
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target = nums[i + 1]
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cnt[target] += 1
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if mx < cnt[target]:
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mx = cnt[target]
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ans = nums[i + 1]
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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 mostFrequent(int[] nums, int key) {
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int[] cnt = new int[1010];
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int mx = 0, ans = 0;
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for (int i = 0; i < nums.length - 1; ++i) {
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if (nums[i] == key) {
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int target = nums[i + 1];
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++cnt[target];
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if (mx < cnt[target]) {
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mx = cnt[target];
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ans = nums[i + 1];
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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 mostFrequent(vector<int>& nums, int key) {
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vector<int> cnt(1010);
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int mx = 0, ans = 0;
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for (int i = 0; i < nums.size() - 1; ++i)
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{
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if (nums[i] == key)
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{
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int target = nums[i + 1];
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++cnt[target];
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if (mx < cnt[target])
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{
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mx = cnt[target];
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ans = nums[i + 1];
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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 mostFrequent(nums []int, key int) int {
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cnt := make([]int, 1010)
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mx, ans := 0, 0
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for i, v := range nums[:len(nums)-1] {
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if v == key {
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target := nums[i+1]
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cnt[target]++
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if mx < cnt[target] {
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mx = cnt[target]
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ans = nums[i+1]
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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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### **TypeScript**

‎solution/2100-2199/2190.Most Frequent Number Following Key In an Array/README_EN.md

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### **Python3**
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```python
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class Solution:
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def mostFrequent(self, nums: List[int], key: int) -> int:
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cnt = Counter()
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mx = ans = 0
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for i, v in enumerate(nums[:-1]):
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if v == key:
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target = nums[i + 1]
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cnt[target] += 1
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if mx < cnt[target]:
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mx = cnt[target]
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ans = nums[i + 1]
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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 mostFrequent(int[] nums, int key) {
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int[] cnt = new int[1010];
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int mx = 0, ans = 0;
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for (int i = 0; i < nums.length - 1; ++i) {
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if (nums[i] == key) {
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int target = nums[i + 1];
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++cnt[target];
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if (mx < cnt[target]) {
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mx = cnt[target];
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ans = nums[i + 1];
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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 mostFrequent(vector<int>& nums, int key) {
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vector<int> cnt(1010);
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int mx = 0, ans = 0;
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for (int i = 0; i < nums.size() - 1; ++i)
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{
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if (nums[i] == key)
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{
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int target = nums[i + 1];
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++cnt[target];
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if (mx < cnt[target])
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{
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mx = cnt[target];
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ans = nums[i + 1];
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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 mostFrequent(nums []int, key int) int {
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cnt := make([]int, 1010)
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mx, ans := 0, 0
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for i, v := range nums[:len(nums)-1] {
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if v == key {
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target := nums[i+1]
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cnt[target]++
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if mx < cnt[target] {
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mx = cnt[target]
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ans = nums[i+1]
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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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### **TypeScript**

‎solution/2100-2199/2190.Most Frequent Number Following Key In an Array/Solution.cpp

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class Solution {
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public:
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int mostFrequent(vector<int>& nums, int key) {
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vector<int> cnt(1010);
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int mx = 0, ans = 0;
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for (int i = 0; i < nums.size() - 1; ++i)
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{
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if (nums[i] == key)
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{
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int target = nums[i + 1];
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++cnt[target];
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if (mx < cnt[target])
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{
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mx = cnt[target];
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ans = nums[i + 1];
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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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};

‎solution/2100-2199/2190.Most Frequent Number Following Key In an Array/Solution.go

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func mostFrequent(nums []int, key int) int {
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cnt := make([]int, 1010)
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mx, ans := 0, 0
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for i, v := range nums[:len(nums)-1] {
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if v == key {
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target := nums[i+1]
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cnt[target]++
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if mx < cnt[target] {
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mx = cnt[target]
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ans = nums[i+1]
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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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