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feat: update leetcode solutions: No.0074. Search a 2D Matrix
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‎README.md

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@@ -50,6 +50,7 @@
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### 数组
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- [删除排序数组中的重复项](/solution/0000-0099/0026.Remove%20Duplicates%20from%20Sorted%20Array/README.md)
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- [删除排序数组中的重复项 II](/solution/0000-0099/0080.Remove%20Duplicates%20from%20Sorted%20Array%20II/README.md)
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- [移除元素](/solution/0000-0099/0027.Remove%20Element/README.md)
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- [移动零](/solution/0200-0299/0283.Move%20Zeroes/README.md)
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- [数组中重复的数字](/lcof/面试题03.%20数组中重复的数字/README.md)

‎README_EN.md

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@@ -50,6 +50,7 @@ Complete solutions to [LeetCode](https://leetcode-cn.com/problemset/all/), [LCOF
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### Arrays
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- [Remove Duplicates from Sorted Array](/solution/0000-0099/0026.Remove%20Duplicates%20from%20Sorted%20Array/README_EN.md)
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- [Remove Duplicates from Sorted Array II](/solution/0000-0099/0080.Remove%20Duplicates%20from%20Sorted%20Array%20II/README_EN.md)
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- [Remove Element](/solution/0000-0099/0027.Remove%20Element/README_EN.md)
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- [Move Zeroes](/solution/0200-0299/0283.Move%20Zeroes/README_EN.md)
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- [Rotate Array](/solution/0100-0199/0189.Rotate%20Array/README_EN.md)

‎solution/0000-0099/0074.Search a 2D Matrix/README.md

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@@ -39,22 +39,70 @@ target = 13
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<!-- 这里可写通用的实现逻辑 -->
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将二维矩阵逻辑展开，然后二分查找即可。
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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 searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
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m, n = len(matrix), len(matrix[0])
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l, h = 0, m * n - 1
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while l <= h:
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mid = l + ((h - l) >> 1)
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x, y = divmod(mid, n)
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if matrix[x][y] == target:
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return True
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if matrix[x][y] < target:
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l = mid + 1
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else:
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h = mid - 1
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return False
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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 boolean searchMatrix(int[][] matrix, int target) {
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int m = matrix.length, n = matrix[0].length;
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int l = 0, h = m * n - 1;
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while (l <= h) {
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int mid = l + ((h - l) >> 1);
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int x = mid / n, y = mid % n;
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if (matrix[x][y] == target) return true;
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if (matrix[x][y] < target) l = mid + 1;
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else h = mid - 1;
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}
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return false;
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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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bool searchMatrix(vector<vector<int>>& matrix, int target) {
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int m = matrix.size(), n = matrix[0].size();
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int l = 0, h = m * n - 1;
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while (l <= h) {
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int mid = l + ((h - l) >> 1);
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int x = mid / n, y = mid % n;
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if (matrix[x][y] == target) return true;
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if (matrix[x][y] < target) l = mid + 1;
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else h = mid - 1;
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}
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return false;
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}
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};
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```
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### **...**

‎solution/0000-0099/0074.Search a 2D Matrix/README_EN.md

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@@ -60,13 +60,59 @@ target = 13
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### **Python3**
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```python
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class Solution:
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def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
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m, n = len(matrix), len(matrix[0])
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l, h = 0, m * n - 1
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while l <= h:
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mid = l + ((h - l) >> 1)
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x, y = divmod(mid, n)
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if matrix[x][y] == target:
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return True
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if matrix[x][y] < target:
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l = mid + 1
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else:
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h = mid - 1
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return False
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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 boolean searchMatrix(int[][] matrix, int target) {
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int m = matrix.length, n = matrix[0].length;
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int l = 0, h = m * n - 1;
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while (l <= h) {
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int mid = l + ((h - l) >> 1);
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int x = mid / n, y = mid % n;
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if (matrix[x][y] == target) return true;
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if (matrix[x][y] < target) l = mid + 1;
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else h = mid - 1;
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}
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return false;
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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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bool searchMatrix(vector<vector<int>>& matrix, int target) {
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int m = matrix.size(), n = matrix[0].size();
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int l = 0, h = m * n - 1;
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while (l <= h) {
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int mid = l + ((h - l) >> 1);
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int x = mid / n, y = mid % n;
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if (matrix[x][y] == target) return true;
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if (matrix[x][y] < target) l = mid + 1;
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else h = mid - 1;
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}
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return false;
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}
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};
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```
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### **...**

‎solution/0000-0099/0074.Search a 2D Matrix/Solution.cpp

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Original file line numberDiff line numberDiff line change
@@ -1,22 +1,15 @@
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class Solution {
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public:
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bool searchMatrix(vector<vector<int>>& matrix, int target) {
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if(matrix.empty())return false;
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size_t row = matrix.size();
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size_t column = matrix[0].size();
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if(column == 0 || column == 0)return false;
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if(target < matrix[0][0] || target > matrix[row-1][column-1])return false;
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for(int i = 0;i<row;i++){
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if(matrix[i][column-1]<target)continue;
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auto iter = find(matrix[i].begin(),matrix[i].end(),target);
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if(iter != matrix[i].end())return true;
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else return false;
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int m = matrix.size(), n = matrix[0].size();
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int l = 0, h = m * n - 1;
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while (l <= h) {
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int mid = l + ((h - l) >> 1);
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int x = mid / n, y = mid % n;
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if (matrix[x][y] == target) return true;
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if (matrix[x][y] < target) l = mid + 1;
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else h = mid - 1;
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}
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return false;
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}
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}
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};

‎solution/0000-0099/0074.Search a 2D Matrix/Solution.java

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@@ -1,13 +1,13 @@
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class Solution {
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public boolean searchMatrix(int[][] matrix, int target) {
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int x = 0,y = matrix.length-1;
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if(y<0) return false;
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int maxX = matrix[0].length-1;
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while ((x <= maxX) && (y >= 0)) {
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int cur = matrix[y][x];
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if (cur == target) return false;
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if (cur < target) x++;
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else y--;
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int m = matrix.length, n = matrix[0].length;
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int l = 0, h = m * n - 1;
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while (l <= h) {
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int mid = l + ((h - l) >> 1);
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int x = mid / n, y = mid % n;
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if (matrix[x][y] == target) return true;
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if (matrix[x][y] < target) l = mid + 1;
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else h = mid - 1;
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}
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return false;
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}

‎solution/0000-0099/0074.Search a 2D Matrix/Solution.py

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@@ -1,27 +1,14 @@
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class Solution:
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def searchMatrix(self, matrix, target):
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"""
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:type matrix: List[List[int]]
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:type target: int
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:rtype: bool
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"""
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if not matrix or not matrix[0]:
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return False
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for row in matrix:
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if row[0] <= target <= row[-1]:
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s = 0
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e = len(row)-1
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while s <= e:
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mid = (s+e)//2
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if row[mid] == target:
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return True
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elif row[mid] > target:
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e = mid-1
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else:
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s = mid+1
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return False
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def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
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m, n = len(matrix), len(matrix[0])
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l, h = 0, m * n - 1
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while l <= h:
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mid = l + ((h - l) >> 1)
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x, y = divmod(mid, n)
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if matrix[x][y] == target:
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return True
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if matrix[x][y] < target:
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l = mid + 1
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else:
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h = mid - 1
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return False

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