Search a 2D Matrix
描述
Write an efficient algorithm that searches for a value in an m × 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.
For example, Consider the following matrix:
[
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
Given target = 3, return true.
分析
二分查找。
代码
- Python
- Java
- C++
// Search a 2D Matrix
// 时间复杂度O(logn)，空间复杂度O(1)
public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if (matrix.length == 0) return false;
final int m = matrix.length;
final int n = matrix[0].length;
int first = 0;
int last = m * n;
while (first < last) {
int mid = first + (last - first) / 2;
int value = matrix[mid / n][mid % n];
if (value == target)
return true;
else if (value < target)
first = mid + 1;
else
last = mid;
}
return false;
}
}
// Search a 2D Matrix
// 时间复杂度O(logn)，空间复杂度O(1)
class Solution {
public:
bool searchMatrix(const vector<vector<int>>& matrix, int target) {
if (matrix.empty()) return false;
const size_t m = matrix.size();
const size_t n = matrix.front().size();
int first = 0;
int last = m * n;
while (first < last) {
int mid = first + (last - first) / 2;
int value = matrix[mid / n][mid % n];
if (value == target)
return true;
else if (value < target)
first = mid + 1;
else
last = mid;
}
return false;
}
};
# Search a 2D Matrix
# 时间复杂度O(logn)，空间复杂度O(1)
def searchMatrix(matrix, target):
if not matrix:
return False
m = len(matrix)
n = len(matrix[0])
first = 0
last = m * n
while first < last:
mid = first + (last - first) // 2
value = matrix[mid // n][mid % n]
if value == target:
return True
elif value < target:
first = mid + 1
else:
last = mid
return False