| # |
Title |
Difficulty |
Topic |
| 74 |
Search a 2D Matrix |
Medium |
Array; Binary Search |
Description
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:
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Input:
matrix = [
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
target = 3
Output: true
|
Example 2:
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Input:
matrix = [
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
target = 13
Output: false
|
Analyze
当作一个排序数组来做。
Submission
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class {
public boolean searchMatrix(int[][] matrix, int target) {
if(matrix == null || matrix.length == 0 ||
matrix[0] == null || matrix[0].length == 0) return false;
int m = matrix.length, n = matrix[0].length;
int lo = 0, hi = m * n - 1;
while(lo < hi) {
int mid = (lo + hi) / 2;
if(matrix[mid / n][mid % n] < target) {
lo = mid + 1;
} else {
hi = mid;
}
}
return matrix[lo / n][lo % n] == target;
}
}
|
Complex Analyze
时间复杂度为$$O(log(m times n))$$。
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