Leetcode 74 Search a 2D Matrix

题目

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

思路

就是一个摊开来的binary search，(x, y)点对应数字x*matrix[0].length + y，由此可以转换为一个长度为matrix.length*matrix[0].length的一维数组。
binary search参考Leetcode 704 Binary Search

代码

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0) return false;
        return search(matrix, 0, matrix.length * matrix[0].length, target);
    }
    
    public boolean search(int[][] matrix, int left, int right, int target) {
        if (left == right) return false;
        int mid = (left + right) / 2;
        if (matrix[mid / matrix[0].length][mid % matrix[0].length] == target) return true;
        if (matrix[mid / matrix[0].length][mid % matrix[0].length] > target) {
            return search(matrix, left, mid, target);
        }
        return search(matrix, mid + 1, right, target);
    }
}