74. 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.

第一种方法：先对第一列用二分法，求出target在哪一列，然后再在列里面用二分法。

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

第二种方法：不要把它看作二维矩阵，把它看成连续的一维数组，再使用二分法，感觉简单很多。

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