leetcode - Search a 2D Matrix II

题目：

Search a 2D Matrix II

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 in ascending from left to right.
• Integers in each column are sorted in ascending from top to bottom.

For example,

Consider the following matrix:

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

Given target = 5, return true.

Given target = 20, return false.

分析：

时间复杂度O(log(min(m,n)))。

for循环每次都仅仅保留矩阵的“左上角”。

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        if(matrix.empty() || matrix[0].empty())
            return false;
        int rows=matrix.size(),cols=matrix[0].size();
        int endrow=cols-1,endcol=rows-1;
        for(int i=0;i<=min(rows,cols)-1;++i)
        {
            int r1=checkrows(matrix,target,i,endrow);
            if(r1==-2)//找到了
                return true;
            if(r1==-1)//所有都大，或者i>endrow
                return false;
            int r2=checkcols(matrix,target,i,endcol);
            if(r2==-2)//找到了
                return true;
            if(r2==-1)//所有都大。或者i>endcol
                return false;
            endrow=r1;
            endcol=r2;
        }
        return false;
    }
    
    int checkrows(vector<vector<int>>& m, int target,int begin,int end)
    {
        int row=begin;
        int res=-1;
        while(begin<=end)
        {
            int mid=begin+(end-begin)/2;
            if(m[row][mid]==target)
                return -2;
            else if(m[row][mid]<target)
            {
                res=mid;
                begin=mid+1;
            }
            else
                end=mid-1;
        }
        return res;
    }
    
     int checkcols(vector<vector<int>>& m, int target,int begin,int end)
    {
        int col=begin;
        int res=-1;
        while(begin<=end)
        {
            int mid=begin+(end-begin)/2;
            if(m[mid][col]==target)
                return -2;
            else if(m[mid][col]<target)
            {
                res=mid;
                begin=mid+1;
            }
            else
                end=mid-1;
        }
        return res;
    }
    
};