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feat: add solutions to lc problem: No.0892
No.0892.Surface Area of 3D Shapes
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‎solution/0800-0899/0892.Surface Area of 3D Shapes/README.md

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@@ -55,22 +55,107 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：遍历，逐个累加**
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时间复杂度 $O(n^2)$，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 surfaceArea(self, grid: List[List[int]]) -> int:
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ans = 0
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for i, row in enumerate(grid):
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for j, v in enumerate(row):
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if v:
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ans += 2 + v * 4
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if i:
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ans -= min(v, grid[i - 1][j]) * 2
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if j:
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ans -= min(v, grid[i][j - 1]) * 2
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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 surfaceArea(int[][] grid) {
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int n = grid.length;
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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 = 0; j < n; ++j) {
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if (grid[i][j] > 0) {
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ans += 2 + grid[i][j] * 4;
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if (i > 0) {
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ans -= Math.min(grid[i][j], grid[i - 1][j]) * 2;
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}
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if (j > 0) {
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ans -= Math.min(grid[i][j], grid[i][j - 1]) * 2;
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}
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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 surfaceArea(vector<vector<int>>& grid) {
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int n = grid.size();
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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 = 0; j < n; ++j) {
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if (grid[i][j]) {
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ans += 2 + grid[i][j] * 4;
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if (i) ans -= min(grid[i][j], grid[i - 1][j]) * 2;
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if (j) ans -= min(grid[i][j], grid[i][j - 1]) * 2;
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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 surfaceArea(grid [][]int) int {
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ans := 0
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for i, row := range grid {
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for j, v := range row {
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if v > 0 {
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ans += 2 + v*4
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if i > 0 {
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ans -= min(v, grid[i-1][j]) * 2
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}
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if j > 0 {
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ans -= min(v, grid[i][j-1]) * 2
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}
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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 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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### **...**

‎solution/0800-0899/0892.Surface Area of 3D Shapes/README_EN.md

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### **Python3**
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```python
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class Solution:
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def surfaceArea(self, grid: List[List[int]]) -> int:
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ans = 0
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for i, row in enumerate(grid):
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for j, v in enumerate(row):
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if v:
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ans += 2 + v * 4
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if i:
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ans -= min(v, grid[i - 1][j]) * 2
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if j:
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ans -= min(v, grid[i][j - 1]) * 2
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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 surfaceArea(int[][] grid) {
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int n = grid.length;
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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 = 0; j < n; ++j) {
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if (grid[i][j] > 0) {
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ans += 2 + grid[i][j] * 4;
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if (i > 0) {
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ans -= Math.min(grid[i][j], grid[i - 1][j]) * 2;
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}
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if (j > 0) {
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ans -= Math.min(grid[i][j], grid[i][j - 1]) * 2;
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}
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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 surfaceArea(vector<vector<int>>& grid) {
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int n = grid.size();
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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 = 0; j < n; ++j) {
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if (grid[i][j]) {
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ans += 2 + grid[i][j] * 4;
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if (i) ans -= min(grid[i][j], grid[i - 1][j]) * 2;
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if (j) ans -= min(grid[i][j], grid[i][j - 1]) * 2;
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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 surfaceArea(grid [][]int) int {
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ans := 0
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for i, row := range grid {
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for j, v := range row {
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if v > 0 {
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ans += 2 + v*4
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if i > 0 {
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ans -= min(v, grid[i-1][j]) * 2
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}
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if j > 0 {
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ans -= min(v, grid[i][j-1]) * 2
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}
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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 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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### **...**

‎solution/0800-0899/0892.Surface Area of 3D Shapes/Solution.cpp

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class Solution {
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public:
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int surfaceArea(vector<vector<int>>& grid) {
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int n = grid.size();
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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 = 0; j < n; ++j) {
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if (grid[i][j]) {
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ans += 2 + grid[i][j] * 4;
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if (i) ans -= min(grid[i][j], grid[i - 1][j]) * 2;
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if (j) ans -= min(grid[i][j], grid[i][j - 1]) * 2;
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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/0800-0899/0892.Surface Area of 3D Shapes/Solution.go

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func surfaceArea(grid [][]int) int {
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ans := 0
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for i, row := range grid {
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for j, v := range row {
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if v > 0 {
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ans += 2 + v*4
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if i > 0 {
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ans -= min(v, grid[i-1][j]) * 2
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}
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if j > 0 {
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ans -= min(v, grid[i][j-1]) * 2
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}
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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 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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}

‎solution/0800-0899/0892.Surface Area of 3D Shapes/Solution.java

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class Solution {
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public int surfaceArea(int[][] grid) {
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int n = grid.length;
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int res = 0;
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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 = 0; j < n; ++j) {
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if (grid[i][j] > 0) {
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res += 2 + grid[i][j] * 4;
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ans += 2 + grid[i][j] * 4;
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if (i > 0) {
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res -= Math.min(grid[i][j], grid[i - 1][j]) * 2;
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ans -= Math.min(grid[i][j], grid[i - 1][j]) * 2;
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}
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if (j > 0) {
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res -= Math.min(grid[i][j], grid[i][j - 1]) * 2;
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ans -= Math.min(grid[i][j], grid[i][j - 1]) * 2;
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}
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}
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}
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}
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return res;
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return ans;
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}
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}
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}

‎solution/0800-0899/0892.Surface Area of 3D Shapes/Solution.py

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class Solution:
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def surfaceArea(self, grid: List[List[int]]) -> int:
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ans = 0
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for i, row in enumerate(grid):
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for j, v in enumerate(row):
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if v:
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ans += 2 + v * 4
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if i:
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ans -= min(v, grid[i - 1][j]) * 2
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if j:
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ans -= min(v, grid[i][j - 1]) * 2
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return ans

‎solution/0800-0899/0894.All Possible Full Binary Trees/README.md

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时间复杂度 $O(2^n)$，空间复杂度 $O(2^n)$。
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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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