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：此题还是比較简单的。主要是用两个二分法，才对第一列使用，再对特定的行使用。
详细代码例如以下：
public class Solution {
public boolean searchMatrix(int[][] m, int target) {
//两个二分搜索
//先搜索第一列，找到确定的行，然后再搜索行
int i = 0;
int j = m.length-1;
int mid = 0;
//搜寻第一列
while(i <= j){
mid = (i + j)/2;
if(m[mid][0] == target){
return true;
}else if(m[mid][0] < target){
i = mid + 1;
}else{
j = mid - 1;
}
}
if(m[mid][0] > target){
if(mid == 0){
return false;
}
mid--;//mid-1
}
//搜寻mid行
i = 0;
j = m[0].length -1;
int k = mid;
//搜寻到了返回true
while(i <= j){
mid = (i + j)/2;
if(m[k][mid] == target){
return true;
}else if(m[k][mid] < target){
i = mid + 1;
}else{
j = mid - 1;
}
}
//没有搜寻到，返回false
return false;
}
}