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：1.二分比较判断可能在哪一行，确定searchRow，再在searchRow中二分找

2. 当成线性表，来二分找。

注意：这两个的复杂度？？？？ 1.O(lgm+lgn) 2.O(lg(m*n))

 

class Solution {public:    bool searchMatrix(vector
     
      
        > &matrix, int target) {        if(matrix.empty()) return false;                int row = matrix.size();        int rowA=0; int rowB = row-1;         int rowMid=0;        int searchRow;        while(rowA
       
        =matrix[rowMid-1][0]){searchRow=rowMid-1;break;}                else rowB=rowMid-2;            }else{                if(target
        
         matrix[searchRow][colB]) return false;        while(colA<=colB){            colMid = (colA+colB)/2;            if(target==matrix[searchRow][colMid]) return true;            if(target