Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
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;
1 public class Solution { 2 public boolean searchMatrix(int[][] matrix, int target) { 3 if(matrix.length == 0 || matrix[0].length == 0) 4 return false; 5 int m = matrix.length; 6 int n = matrix[0].length; 7 8 int start = 0,end = m*n-1; 9 while(start <= end){ 10 int mid = start + (end-start)/2; 11 int x = mid/n; 12 int y = mid%n; 13 14 if(matrix[x][y] == target) 15 return true; 16 17 if(matrix[x][y] < target) 18 start = mid+1; 19 else 20 end = mid-1; 21 } 22 return false; 23 } 24 }
【leetcode】Search a 2D Matrix,布布扣,bubuko.com
原文：http://www.cnblogs.com/sunshineatnoon/p/3798224.html