题目：
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 in ascending from left to right.
- Integers in each column are sorted in ascending from top to bottom.
For example,
Consider the following matrix:
[ [1, 4, 7, 11, 15], [2, 5, 8, 12, 19], [3, 6, 9, 16, 22], [10, 13, 14, 17, 24], [18, 21, 23, 26, 30] ]
Given target = 5, return true.
Given target = 20, return false.
题意：给出一个m*n的二维矩阵，具有以下特点：
(1)每一行按照升序排序
(2)每一列按照升序排列
现给出一个值，求该值是否在二维矩阵中存在。这里直接二次循环查找时间复杂度为o(m*n)，肯定会超时的。所以我先查找可能出现target的列和行，然后对可能的行和列分别做二分查找。
代码：
public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if(matrix.length==0||matrix[0].length==0) //空矩阵返回false
return false;
int row=matrix.length;
int col=matrix[0].length;
int mayRow=-1; //可能的行
int mayCol=-1; //可能的列
for(int i=0;i<row;i++){ //寻找可能出现target的行
if(matrix[i][0]<=target&&matrix[i][col-1]>=target){
mayRow=i;
int low=0;
int high=col-1;
while(low<=high){
int mid=(low+high)/2;
if(matrix[mayRow][mid]==target){
System.out.println("true");
return true;
}else if(matrix[mayRow][mid]>target){
high=mid-1;
}else if(matrix[mayRow][mid]<target){
low=mid+1;
}
}
}
}
for(int i=0;i<col;i++){ //寻找可能出现target的列
if(matrix[0][i]<=target&&matrix[row-1][i]>=target){
mayCol=i;
int low=0;
int high=row-1;
while(low<=high){
int mid=(low+high)/2;
if(matrix[mid][mayCol]==target){
System.out.println("true");
return true;
}else if(matrix[mid][mayCol]>target){
high=mid-1;
}else if(matrix[mid][mayCol]<target){
low=mid+1;
}
}
}
}
System.out.println("false");
return false;
}
}