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题目：
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.
解题思路：利用两次二分查找法。因为所给矩阵第一列也是升序排列的，所以可以先对第一列进行二分查找，锁定该元素所在行数，然后再对列进行二分查找，即可判断target是否存在。这个的算法时间复杂度是O(log(rows)+log(columns))。
代码：
1 public class Solution { 2 public boolean searchMatrix(int[][] matrix, int target) 3 { 4 if(matrix.length==0||matrix[0].length==0||matrix==null) return false; 5 6 int low=0; 7 int high=matrix.length-1; 8 9 while(low<=high) 10 { 11 int mid=(high+low)/2; 12 if(target>matrix[mid][0]) 13 { 14 low=mid+1; 15 } 16 else if (target<matrix[mid][0]) 17 { 18 high=mid-1; 19 } 20 else return true; 21 } 22 23 int row=high; //当从while中跳出时，low指向的值肯定比target大，而high指向的值肯定比target小 24 25 if(row<0) return false; 26 27 low=0; 28 high=matrix[0].length-1; 29 while(low<=high) 30 { 31 int mid=(high+low)/2; 32 if(target>matrix[row][mid]) 33 { 34 low=mid+1; 35 } 36 else if (target<matrix[row][mid]) 37 { 38 high=mid-1; 39 } 40 else return true; 41 } 42 43 return false; 44 } 45 }
reference： http://blog.csdn.net/linhuanmars/article/details/24216235
http://www.cnblogs.com/springfor/p/3857959.html
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原文地址：http://www.cnblogs.com/hygeia/p/4651751.html
