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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