Search a 2D Matrix - Leetcode

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

分析：二叉搜索，关键是根据Mid还原在二维数组里面的位置，找到值，与target比较。

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.