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：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<vector<int> > &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<rowB){
            rowMid = (rowA+rowB)/2;
            if(target<matrix[rowMid][0]){
                if(rowMid-1<0) return false;
                else if(target>=matrix[rowMid-1][0]){searchRow=rowMid-1;break;}
                else rowB=rowMid-2;
            }else{
                if(target<matrix[rowMid+1][0]){searchRow=rowMid;break;}
                else rowA=rowMid+1;
            }
        }
        if(rowA==rowB) searchRow = rowA;
        
        int col = matrix[0].size();
        int colA=0; int colB=col-1;
        int colMid=0;
        if(target<matrix[searchRow][colA] || target>matrix[searchRow][colB]) return false;
        while(colA<=colB){
            colMid = (colA+colB)/2;
            if(target==matrix[searchRow][colMid]) return true;
            if(target<matrix[searchRow][colMid]) colB=colMid-1;
            else colA=colMid+1;
        }
        return false;
    }
};