Write an efficient algorithm that searches for a value in an m x n
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.
Example 1:
Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,50]], target = 3 Output: true
Example 2:
Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,50]], target = 13 Output: false
Example 3:
Input: matrix = [], target = 0 Output: false
Constraints:
m == matrix.length
n == matrix[i].length
0 <= m, n <= 100
-104 <= matrix[i][j], target <= 104
Related Topics:
Array, Binary Search
Similar Questions:
// OJ: https://leetcode.com/problems/search-a-2d-matrix/
// Author: github.com/lzl124631x
// Time: O(log(MN))
// Space: O(1)
class Solution {
public:
bool searchMatrix(vector<vector<int>>& A, int target) {
if (A.empty() || A[0].empty()) return false;
int L = 0, R = A.size() - 1;
while (L <= R) {
int M = (L + R) / 2;
if (A[M][0] == target) return true;
if (A[M][0] < target) L = M + 1;
else R = M - 1;
}
if (R == -1) return false;
int row = R;
L = 0, R = A[0].size() - 1;
while (L <= R) {
int M = (L + R) / 2;
if (A[row][M] == target) return true;
if (A[row][M] < target) L = M + 1;
else R = M - 1;
}
return false;
}
};
// OJ: https://leetcode.com/problems/search-a-2d-matrix/
// Author: github.com/lzl124631x
// Time: O(log(MN))
// Space: O(1)
class Solution {
public:
bool searchMatrix(vector<vector<int>>& A, int target) {
if (A.empty() || A[0].empty()) return false;
int M = A.size(), N = A[0].size(), L = 0, R = M * N - 1;
while (L <= R) {
int mid = (L + R) / 2, val = A[mid / N][mid % N];
if (val == target) return true;
if (val < target) L = mid + 1;
else R = mid - 1;
}
return false;
}
};
// OJ: https://leetcode.com/problems/search-a-2d-matrix/
// Author: github.com/lzl124631x
// Time: O(M + N)
// Space: O(1)
class Solution {
public:
bool searchMatrix(vector<vector<int>>& A, int target) {
if (A.empty() || A[0].empty()) return false;
int M = A.size(), N = A[0].size(), i = 0, j = N - 1;
while (i < M && j >= 0) {
if (A[i][j] == target) return true;
if (A[i][j] < target) ++i;
else --j;
}
return false;
}
};