74. Search a 2D Matrix

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 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

本题矩阵的各种限制条件就是说明这个矩阵是有序的，查找target就是用二分法：

class Solution {public:    bool searchMatrix(vector<vector<int>>& matrix, int target) {         if(matrix.empty())            return false;        int m=matrix.size(),n=matrix[0].size();        int first =0,end =m*n-1,mid = (m*n-1) / 2 ;        while(first<=end)        {            if(target>matrix[m-1][n-1]||target<matrix[0][0])                return false;            else if(target>matrix[mid/n][mid-(mid/n)*n])            {                first = mid+1;                mid = (first+end) /2;            }            else if(target<matrix[mid/n][mid-(mid/n)*n])            {                end = mid - 1;                mid = (first+end) /2;            }            else if(target==matrix[mid/n][mid-(mid/n)*n])                return true;        }        return false;        }};