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feat: add new lc problems and solutions
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‎solution/0500-0599/0532.K-diff Pairs in an Array/README.md

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Original file line numberDiff line numberDiff line change
@@ -58,22 +58,92 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：哈希表**
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由于 $k$ 是一个定值，因此用哈希表 $ans$ 记录数对的较小值，就能够确定较大的值。最后返回 ans 的大小作为答案。
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遍历数组 $nums$，当前遍历到的数 $nums[j]$，我们记为 $v$，用哈希表 $vis$ 记录此前遍历到的所有数字。若 $v-k$ 在 $vis$ 中，则将 $v-k$ 添加至 $ans$；若 $v+k$ 在 $vis$ 中，则将 $v$ 添加至 $ans$。
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时间复杂度 $O(n)$，其中 $n$ 表示数组 $nums$ 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 findPairs(self, nums: List[int], k: int) -> int:
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vis, ans = set(), set()
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for v in nums:
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if v - k in vis:
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ans.add(v - k)
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if v + k in vis:
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ans.add(v)
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vis.add(v)
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return len(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 findPairs(int[] nums, int k) {
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Set<Integer> vis = new HashSet<>();
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Set<Integer> ans = new HashSet<>();
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for (int v : nums) {
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if (vis.contains(v - k)) {
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ans.add(v - k);
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}
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if (vis.contains(v + k)) {
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ans.add(v);
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}
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vis.add(v);
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}
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return ans.size();
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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 findPairs(vector<int>& nums, int k) {
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unordered_set<int> vis;
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unordered_set<int> ans;
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for (int& v : nums)
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{
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if (vis.count(v - k)) ans.insert(v - k);
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if (vis.count(v + k)) ans.insert(v);
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vis.insert(v);
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}
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return ans.size();
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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 findPairs(nums []int, k int) int {
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vis := map[int]bool{}
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ans := map[int]bool{}
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for _, v := range nums {
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if vis[v-k] {
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ans[v-k] = true
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}
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if vis[v+k] {
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ans[v] = true
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}
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vis[v] = true
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}
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return len(ans)
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}
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```
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### **...**

‎solution/0500-0599/0532.K-diff Pairs in an Array/README_EN.md

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Original file line numberDiff line numberDiff line change
@@ -58,13 +58,75 @@ Although we have two 1s in the input, we should only return the number of <stron
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### **Python3**
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```python
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class Solution:
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def findPairs(self, nums: List[int], k: int) -> int:
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vis, ans = set(), set()
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for v in nums:
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if v - k in vis:
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ans.add(v - k)
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if v + k in vis:
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ans.add(v)
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vis.add(v)
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return len(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 findPairs(int[] nums, int k) {
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Set<Integer> vis = new HashSet<>();
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Set<Integer> ans = new HashSet<>();
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for (int v : nums) {
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if (vis.contains(v - k)) {
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ans.add(v - k);
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}
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if (vis.contains(v + k)) {
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ans.add(v);
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}
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vis.add(v);
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}
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return ans.size();
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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 findPairs(vector<int>& nums, int k) {
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unordered_set<int> vis;
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unordered_set<int> ans;
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for (int& v : nums)
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{
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if (vis.count(v - k)) ans.insert(v - k);
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if (vis.count(v + k)) ans.insert(v);
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vis.insert(v);
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}
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return ans.size();
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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 findPairs(nums []int, k int) int {
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vis := map[int]bool{}
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ans := map[int]bool{}
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for _, v := range nums {
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if vis[v-k] {
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ans[v-k] = true
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}
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if vis[v+k] {
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ans[v] = true
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}
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vis[v] = true
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}
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return len(ans)
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}
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```
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### **...**

‎solution/0500-0599/0532.K-diff Pairs in an Array/Solution.cpp

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class Solution {
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public:
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int findPairs(vector<int>& nums, int k) {
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unordered_set<int> vis;
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unordered_set<int> ans;
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for (int& v : nums)
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{
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if (vis.count(v - k)) ans.insert(v - k);
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if (vis.count(v + k)) ans.insert(v);
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vis.insert(v);
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}
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return ans.size();
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}
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};

‎solution/0500-0599/0532.K-diff Pairs in an Array/Solution.go

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func findPairs(nums []int, k int) int {
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vis := map[int]bool{}
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ans := map[int]bool{}
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for _, v := range nums {
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if vis[v-k] {
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ans[v-k] = true
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}
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if vis[v+k] {
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ans[v] = true
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}
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vis[v] = true
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}
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return len(ans)
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}

‎solution/0500-0599/0532.K-diff Pairs in an Array/Solution.java

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class Solution {
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public int findPairs(int[] nums, int k) {
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Set<Integer> vis = new HashSet<>();
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Set<Integer> ans = new HashSet<>();
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for (int v : nums) {
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if (vis.contains(v - k)) {
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ans.add(v - k);
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}
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if (vis.contains(v + k)) {
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ans.add(v);
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}
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vis.add(v);
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}
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return ans.size();
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}
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}

‎solution/0500-0599/0532.K-diff Pairs in an Array/Solution.py

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class Solution:
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def findPairs(self, nums: List[int], k: int) -> int:
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vis, ans = set(), set()
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for v in nums:
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if v - k in vis:
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ans.add(v - k)
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if v + k in vis:
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ans.add(v)
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vis.add(v)
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return len(ans)

‎solution/2000-2099/2017.Grid Game/README.md

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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 gridGame(self, grid: List[List[int]]) -> int:
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ans = inf
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s1, s2 = sum(grid[0]), 0
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for j, v in enumerate(grid[0]):
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s1 -= v
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ans = min(ans, max(s1, s2))
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s2 += grid[1][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 long gridGame(int[][] grid) {
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long ans = Long.MAX_VALUE;
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long s1 = 0, s2 = 0;
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for (int v : grid[0]) {
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s1 += v;
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}
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int n = grid[0].length;
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for (int j = 0; j < n; ++j) {
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s1 -= grid[0][j];
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ans = Math.min(ans, Math.max(s1, s2));
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s2 += grid[1][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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using ll = long long;
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class Solution {
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public:
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long long gridGame(vector<vector<int>>& grid) {
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ll ans = LONG_MAX;
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int n = grid[0].size();
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ll s1 = 0, s2 = 0;
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for (int& v : grid[0]) s1 += v;
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for (int j = 0; j < n; ++j)
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{
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s1 -= grid[0][j];
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ans = min(ans, max(s1, s2));
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s2 += grid[1][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 gridGame(grid [][]int) int64 {
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ans := math.MaxInt64
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s1, s2 := 0, 0
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for _, v := range grid[0] {
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s1 += v
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}
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for j, v := range grid[0] {
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s1 -= v
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ans = min(ans, max(s1, s2))
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s2 += grid[1][j]
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}
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return int64(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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func min(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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