标签：
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
For example,
Consider the following matrix:
[ [1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50] ]
Given target = 3, return true.
class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
bool flag=false;
if(matrix.size()==0) return flag;
int m=matrix.size();
int n=matrix[0].size();
int row=0;
for(int i=0;i<m;i++)
{
if(target<matrix[i][0])
{
if(i==0)
return flag;
else
row=i-1;
break;
}
else if(target==matrix[i][0]||target==matrix[i][n-1])
{
return true;
}
else if(target>matrix[i][0]&&target<matrix[i][n-1])
{
row=i;
break;
}
}
if(target>matrix[row][n-1])
return flag;
else if(target==matrix[row][n-1])
{
return true;
}
int low=0,high=n-1;
while(low<=high)
{
int mid=(low+high)/2;
if(target<matrix[row][mid])
high=mid-1;
else if(target>matrix[row][mid])
low=mid+1;
else
{
flag=1;
break;
}
}
return flag;
}
};class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
bool flag=false;
if(matrix.size()==0) return flag;
int m=matrix.size();
int n=matrix[0].size();
if(m>0&&n>0)
{
int row=0;
int col=n-1;
while(row<m&&col>=0)
{
if(matrix[row][col]==target)
{
flag=1;
break;
}
else if(matrix[row][col]>target)
col--;
else
row++;
}
}
return flag;
}
};标签：
原文地址：http://blog.csdn.net/sinat_24520925/article/details/46475883
