【leetcode】Search a 2D Matrix

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

---

题解：直接把二维数组看成一个长度为m*n的数列，然后用二分法查找target，注意在这个数列中编号为k的数在二维数组中的坐标x = k/n, y=k%n;

代码如下;

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix.length == 0 || matrix[0].length == 0)
            return false;
        int m = matrix.length;
        int n = matrix[0].length;
        
        int start = 0,end = m*n-1;
        while(start <= end){
            int mid = start + (end-start)/2;
            int x = mid/n;
            int y = mid%n;
            
            if(matrix[x][y] == target)
                return true;
            
            if(matrix[x][y] < target)
                start = mid+1;
            else
                end = mid-1;
        }
        return false;
    }
}