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feat: add solutions to lc problem: No.1762
No.1762.Buildings With an Ocean View
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‎solution/1700-1799/1762.Buildings With an Ocean View/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：逆序遍历求右侧最大值**
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逆序遍历数组 $height$ 每个元素 $v$，判断 $v$ 与右侧最大元素 $mx$ 的大小关系，若 $mx \lt v$，说明右侧所有元素都比当前元素小，当前位置能看到海景，加入结果数组 $ans$。
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最后逆序返回 $ans$。
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时间复杂度 $O(n)$，空间复杂度 $O(1)$。
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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 findBuildings(self, heights: List[int]) -> List[int]:
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mx = 0
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ans = []
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for i in range(len(heights) - 1, -1, -1):
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v = heights[i]
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if mx < v:
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ans.append(i)
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mx = v
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return ans[::-1]
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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[] findBuildings(int[] heights) {
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int mx = 0;
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LinkedList<Integer> ans = new LinkedList<>();
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for (int i = heights.length - 1; i >= 0; --i) {
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int v = heights[i];
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if (mx < v) {
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ans.addFirst(i);
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mx = v;
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}
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}
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return ans.stream().mapToInt(i -> i).toArray();
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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<int> findBuildings(vector<int>& heights) {
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int mx = 0;
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vector<int> ans;
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for (int i = heights.size() - 1; ~i; --i) {
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int v = heights[i];
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if (mx < v) {
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ans.push_back(i);
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mx = v;
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}
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}
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reverse(ans.begin(), ans.end());
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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 findBuildings(heights []int) []int {
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mx := 0
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ans := []int{}
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for i := len(heights) - 1; i >= 0; i-- {
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v := heights[i]
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if mx < v {
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ans = append(ans, i)
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mx = v
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}
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}
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for i, j := 0, len(ans)-1; i < j; i, j = i+1, j-1 {
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ans[i], ans[j] = ans[j], ans[i]
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}
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return ans
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}
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```
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### **JavaScript**
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```js
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/**
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* @param {number[]} heights
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* @return {number[]}
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*/
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var findBuildings = function (heights) {
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let mx = 0;
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let ans = [];
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for (let i = heights.length - 1; i >= 0; --i) {
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const v = heights[i];
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if (mx < v) {
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ans.push(i);
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mx = v;
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}
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}
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return ans.reverse();
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};
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```
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### **...**

‎solution/1700-1799/1762.Buildings With an Ocean View/README_EN.md

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### **Python3**
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```python
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class Solution:
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def findBuildings(self, heights: List[int]) -> List[int]:
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mx = 0
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ans = []
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for i in range(len(heights) - 1, -1, -1):
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v = heights[i]
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if mx < v:
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ans.append(i)
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mx = v
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return ans[::-1]
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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[] findBuildings(int[] heights) {
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int mx = 0;
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LinkedList<Integer> ans = new LinkedList<>();
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for (int i = heights.length - 1; i >= 0; --i) {
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int v = heights[i];
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if (mx < v) {
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ans.addFirst(i);
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mx = v;
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}
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}
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return ans.stream().mapToInt(i -> i).toArray();
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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<int> findBuildings(vector<int>& heights) {
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int mx = 0;
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vector<int> ans;
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for (int i = heights.size() - 1; ~i; --i) {
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int v = heights[i];
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if (mx < v) {
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ans.push_back(i);
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mx = v;
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}
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}
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reverse(ans.begin(), ans.end());
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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 findBuildings(heights []int) []int {
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mx := 0
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ans := []int{}
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for i := len(heights) - 1; i >= 0; i-- {
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v := heights[i]
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if mx < v {
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ans = append(ans, i)
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mx = v
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}
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}
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for i, j := 0, len(ans)-1; i < j; i, j = i+1, j-1 {
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ans[i], ans[j] = ans[j], ans[i]
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}
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return ans
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}
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```
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### **JavaScript**
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```js
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/**
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* @param {number[]} heights
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* @return {number[]}
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*/
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var findBuildings = function (heights) {
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let mx = 0;
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let ans = [];
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for (let i = heights.length - 1; i >= 0; --i) {
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const v = heights[i];
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if (mx < v) {
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ans.push(i);
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mx = v;
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}
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}
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return ans.reverse();
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};
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```
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### **...**

‎solution/1700-1799/1762.Buildings With an Ocean View/Solution.cpp

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class Solution {
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public:
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vector<int> findBuildings(vector<int>& heights) {
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int mx = 0;
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vector<int> ans;
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for (int i = heights.size() - 1; ~i; --i) {
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int v = heights[i];
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if (mx < v) {
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ans.push_back(i);
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mx = v;
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}
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}
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reverse(ans.begin(), ans.end());
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return ans;
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}
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};

‎solution/1700-1799/1762.Buildings With an Ocean View/Solution.go

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func findBuildings(heights []int) []int {
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mx := 0
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ans := []int{}
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for i := len(heights) - 1; i >= 0; i-- {
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v := heights[i]
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if mx < v {
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ans = append(ans, i)
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mx = v
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}
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}
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for i, j := 0, len(ans)-1; i < j; i, j = i+1, j-1 {
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ans[i], ans[j] = ans[j], ans[i]
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}
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return ans
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}

‎solution/1700-1799/1762.Buildings With an Ocean View/Solution.java

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class Solution {
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public int[] findBuildings(int[] heights) {
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int mx = 0;
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LinkedList<Integer> ans = new LinkedList<>();
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for (int i = heights.length - 1; i >= 0; --i) {
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int v = heights[i];
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if (mx < v) {
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ans.addFirst(i);
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mx = v;
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}
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}
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return ans.stream().mapToInt(i -> i).toArray();
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}
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}

‎solution/1700-1799/1762.Buildings With an Ocean View/Solution.js

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/**
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* @param {number[]} heights
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* @return {number[]}
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*/
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var findBuildings = function (heights) {
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let mx = 0;
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let ans = [];
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for (let i = heights.length - 1; i >= 0; --i) {
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const v = heights[i];
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if (mx < v) {
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ans.push(i);
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mx = v;
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}
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}
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return ans.reverse();
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};

‎solution/1700-1799/1762.Buildings With an Ocean View/Solution.py

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class Solution:
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def findBuildings(self, heights: List[int]) -> List[int]:
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mx = 0
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ans = []
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for i in range(len(heights) - 1, -1, -1):
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v = heights[i]
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if mx < v:
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ans.append(i)
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mx = v
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return ans[::-1]

‎solution/2300-2399/2371.Minimize Maximum Value in a Grid/README.md

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# [2371. Minimize Maximum Value in a Grid](https://leetcode.cn/problems/minimize-maximum-value-in-a-grid)
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[English Version](/solution/2300-2399/2371.Minimize%20Maximum%20Value%20in%20a%20Grid/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>You are given an <code>m x n</code> integer matrix <code>grid</code> containing <strong>distinct</strong> positive integers.</p>
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<p>You have to replace each integer in the matrix with a positive integer satisfying the following conditions:</p>
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<ul>
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<li>The <strong>relative</strong> order of every two elements that are in the same row or column should stay the <strong>same</strong> after the replacements.</li>
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<li>The <strong>maximum</strong> number in the matrix after the replacements should be as <strong>small</strong> as possible.</li>
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</ul>
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<p>The relative order stays the same if for all pairs of elements in the original matrix such that <code>grid[r<sub>1</sub>][c<sub>1</sub>] &gt; grid[r<sub>2</sub>][c<sub>2</sub>]</code> where either <code>r<sub>1</sub> == r<sub>2</sub></code> or <code>c<sub>1</sub> == c<sub>2</sub></code>, then it must be true that <code>grid[r<sub>1</sub>][c<sub>1</sub>] &gt; grid[r<sub>2</sub>][c<sub>2</sub>]</code> after the replacements.</p>
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<p>For example, if <code>grid = [[2, 4, 5], [7, 3, 9]]</code> then a good replacement could be either <code>grid = [[1, 2, 3], [2, 1, 4]]</code> or <code>grid = [[1, 2, 3], [3, 1, 4]]</code>.</p>
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<p>Return <em>the <strong>resulting</strong> matrix.</em> If there are multiple answers, return <strong>any</strong> of them.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2022/08/09/grid2drawio.png" style="width: 371px; height: 121px;" />
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<pre>
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<strong>Input:</strong> grid = [[3,1],[2,5]]
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<strong>Output:</strong> [[2,1],[1,2]]
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<strong>Explanation:</strong> The above diagram shows a valid replacement.
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The maximum number in the matrix is 2. It can be shown that no smaller value can be obtained.
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> grid = [[10]]
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<strong>Output:</strong> [[1]]
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<strong>Explanation:</strong> We replace the only number in the matrix with 1.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>m == grid.length</code></li>
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<li><code>n == grid[i].length</code></li>
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<li><code>1 &lt;= m, n &lt;= 1000</code></li>
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<li><code>1 &lt;= m * n &lt;= 10<sup>5</sup></code></li>
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<li><code>1 &lt;= grid[i][j] &lt;= 10<sup>9</sup></code></li>
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<li><code>grid</code> consists of distinct integers.</li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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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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```
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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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```
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### **TypeScript**
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```ts
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```
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### **...**
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```
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```
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<!-- tabs:end -->

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