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feat: add solutions to lc problem: No.0302
No.0302.Smallest Rectangle Enclosing Black Pixels
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‎solution/0300-0399/0302.Smallest Rectangle Enclosing Black Pixels/README.md

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<!-- 这里可写通用的实现逻辑 -->
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二分查找，时间复杂度 `O(mlogn + nlogm)`。
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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 minArea(self, image: List[List[str]], x: int, y: int) -> int:
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m, n = len(image), len(image[0])
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left, right = 0, x
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while left < right:
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mid = (left + right) >> 1
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c = 0
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while c < n and image[mid][c] == '0':
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c += 1
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if c < n:
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right = mid
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else:
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left = mid + 1
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u = left
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left, right = x, m - 1
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while left < right:
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mid = (left + right + 1) >> 1
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c = 0
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while c < n and image[mid][c] == '0':
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c += 1
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if c < n:
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left = mid
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else:
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right = mid - 1
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d = left
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left, right = 0, y
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while left < right:
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mid = (left + right) >> 1
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r = 0
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while r < m and image[r][mid] == '0':
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r += 1
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if r < m:
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right = mid
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else:
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left = mid + 1
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l = left
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left, right = y, n - 1
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while left < right:
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mid = (left + right + 1) >> 1
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r = 0
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while r < m and image[r][mid] == '0':
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r += 1
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if r < m:
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left = mid
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else:
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right = mid - 1
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r = left
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return (d - u + 1) * (r - l + 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 minArea(char[][] image, int x, int y) {
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int m = image.length, n = image[0].length;
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int left = 0, right = x;
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while (left < right) {
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int mid = (left + right) >> 1;
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int c = 0;
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while (c < n && image[mid][c] == '0') {
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++c;
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}
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if (c < n) {
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right = mid;
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} else {
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left = mid + 1;
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}
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}
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int u = left;
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left = x;
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right = m - 1;
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while (left < right) {
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int mid = (left + right + 1) >> 1;
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int c = 0;
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while (c < n && image[mid][c] == '0') {
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++c;
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}
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if (c < n) {
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left = mid;
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} else {
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right = mid - 1;
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}
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}
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int d = left;
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left = 0;
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right = y;
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while (left < right) {
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int mid = (left + right) >> 1;
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int r = 0;
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while (r < m && image[r][mid] == '0') {
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++r;
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}
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if (r < m) {
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right = mid;
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} else {
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left = mid + 1;
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}
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}
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int l = left;
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left = y;
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right = n - 1;
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while (left < right) {
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int mid = (left + right + 1) >> 1;
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int r = 0;
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while (r < m && image[r][mid] == '0') {
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++r;
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}
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if (r < m) {
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left = mid;
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} else {
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right = mid - 1;
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}
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}
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int r = left;
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return (d - u + 1) * (r - l + 1);
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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 minArea(vector<vector<char>>& image, int x, int y) {
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int m = image.size(), n = image[0].size();
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int left = 0, right = x;
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while (left < right)
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{
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int mid = (left + right) >> 1;
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int c = 0;
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while (c < n && image[mid][c] == '0') ++c;
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if (c < n) right = mid;
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else left = mid + 1;
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}
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int u = left;
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left = x;
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right = m - 1;
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while (left < right)
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{
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int mid = (left + right + 1) >> 1;
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int c = 0;
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while (c < n && image[mid][c] == '0') ++c;
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if (c < n) left = mid;
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else right = mid - 1;
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}
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int d = left;
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left = 0;
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right = y;
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while (left < right)
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{
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int mid = (left + right) >> 1;
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int r = 0;
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while (r < m && image[r][mid] == '0') ++r;
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if (r < m) right = mid;
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else left = mid + 1;
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}
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int l = left;
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left = y;
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right = n - 1;
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while (left < right)
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{
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int mid = (left + right + 1) >> 1;
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int r = 0;
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while (r < m && image[r][mid] == '0') ++r;
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if (r < m) left = mid;
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else right = mid - 1;
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}
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int r = left;
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return (d - u + 1) * (r - l + 1);
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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 minArea(image [][]byte, x int, y int) int {
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m, n := len(image), len(image[0])
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left, right := 0, x
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for left < right {
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mid := (left + right) >> 1
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c := 0
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for c < n && image[mid][c] == '0' {
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c++
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}
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if c < n {
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right = mid
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} else {
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left = mid + 1
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}
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}
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u := left
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left, right = x, m-1
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for left < right {
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mid := (left + right + 1) >> 1
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c := 0
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for c < n && image[mid][c] == '0' {
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c++
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}
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if c < n {
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left = mid
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} else {
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right = mid - 1
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}
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}
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d := left
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left, right = 0, y
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for left < right {
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mid := (left + right) >> 1
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r := 0
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for r < m && image[r][mid] == '0' {
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r++
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}
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if r < m {
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right = mid
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} else {
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left = mid + 1
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}
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}
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l := left
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left, right = y, n-1
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for left < right {
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mid := (left + right + 1) >> 1
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r := 0
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for r < m && image[r][mid] == '0' {
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r++
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}
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if r < m {
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left = mid
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} else {
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right = mid - 1
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}
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}
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r := left
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return (d - u + 1) * (r - l + 1)
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}
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```
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### **...**

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