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.

思路：利用matrix结构，倒着搜索，O(m+n);

public class Solution {     public boolean searchMatrix(int[][] matrix, int target) {         if(matrix == null || matrix[0].length == 0) return false;         int m = matrix.length;          int n = matrix[0].length;         int x = 0;          int y = matrix[0].length-1;              while(y>=0 && x<m){             if(matrix[x][y]<target){                 x++;             } else if(matrix[x][y] > target){                 y--;             } else {                 return true;             }         }         return false;     } }

但是这题最优解法是binary search:

思路： 用binary search，将二维拉伸成一维，然后动态算坐标。

public class Solution {     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<=end){             int mid = start +(end-start)/2;             int x = mid/n;             int y = mid%n;             if(matrix[x][y] > target){                 end = mid-1;             } else if(matrix[x][y] < target){                 start = mid+1;             } else {                 return true;             }         }         return false;     } }