Write an efficient algorithm that searches for a value target in an m x n integer matrix. This matrix has the following properties:
- Integers in each row are sorted from left to right.
- The first integer of each row is greater than the last integer of the previous row.
Test Cases
Example 1:
| 1 | 3 | 5 | 7 |
| 10 | 11 | 16 | 20 |
| 23 | 30 | 34 | 60 |
Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
Output: true
Example 2:
| 1 | 3 | 5 | 7 |
| 10 | 11 | 16 | 20 |
| 23 | 30 | 34 | 60 |
Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
Output: false
Solution
bool searchMatrix(int** matrix, int matrixSize, int* matrixColSize, int target){
int row = 0, col = *matrixColSize-1;
while(row < matrixSize && col >= 0) {
int cell = matrix[row][col];
if (cell == target) return true;
if (cell < target) {
row++;
} else {
col--;
}
}
return false;
}class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
int n = matrix.size(), m = matrix[0].size();
int row = 0, col = m-1;
while(row < n && col >= 0) {
int cell = matrix[row][col];
if (cell == target) return true;
if (cell < target) {
row++;
} else {
col--;
}
}
return false;
}
};package search_a_2d_matrix
func searchMatrix(matrix [][]int, target int) bool {
n := len(matrix)
m := len(matrix[0])
row := 0
col := m-1
for row < n && col >=0 {
cell := matrix[row][col]
if cell == target {
return true
}
if cell < target {
row++
} else {
col--
}
}
return false
}class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
int n = matrix.length, m = matrix[0].length;
int row = 0, col = m-1;
while(row < n && col >= 0) {
int cell = matrix[row][col];
if (cell == target) return true;
if (cell < target) {
row++;
} else {
col--;
}
}
return false;
}
}from typing import List
class Solution:
def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
n, m = len(matrix), len(matrix[0])
row, col = 0, m - 1
while row < n and col >= 0:
cell = matrix[row][col]
if cell == target:
return True
if cell < target:
row += 1
else:
col -= 1
return False