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.
For example,
Consider the following matrix:
[ [1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50] ]
Given target = 3, return true.
二分法。
public class Solution {
//Time: O(log(m*n)) Space: O(1)
public boolean searchMatrix(int[][] matrix, int target) {
if (matrix == null || matrix.length == 0) {
return false;
}
int m = matrix.length;
int n = matrix[0].length;
int start = 0;
int end = m * n - 1;
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (matrix[mid / n][mid % n] == target) {
return true;
} else if (matrix[mid / n][mid % n] > target) {
end = mid;
} else {
start = mid;
}
}
if (matrix[start / n][start % n] == target || matrix[end / n][end % n] == target) {
return true;
}
return false;
}
}
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