Leetcode——240. Search a 2D Matrix II

description

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.

https://leetcode.com/problems/search-a-2d-matrix-ii/?tab=Description

solution

第一种，遍历每一行，或者每一列（具体要根据行数列数少的那个进行遍历），对于每一行（每一列）进行二分查找算法为O(mlog(n))或者O(nlog(m))

class Solution1 {//O(Mlog(N)
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
int m=matrix.size();
if(m==0) return false;
int n=matrix[0].size();
if(n==0) return false;
if(m<=n)
{
for(int i=0;i<m;i++)
{
int left=0,right=n-1;
int mid=(left+right)/2;
while(left<=right)
{
// if(right-left==1) return matrix[i][left]==target||matrix[i][right]==target;
if(matrix[i][mid]==target)
return true;
else if(matrix[i][mid]>target)
{
right=mid-1;
mid=(left+right)/2;
}
else
{
left=mid+1;
mid=(left+right)/2;
}
}

}
}
else
{
for(int i=0;i<n;i++)
{
int left=0,right=m-1;
int mid=(left+right)/2;
while(left<=right)
{
// if(right-left==1) return matrix[left][i]==target||matrix[right][i]==target;
if(matrix[mid][i]==target)
return true;
else if(matrix[mid][i]>target)
{
right=mid-1;
mid=(left+right)/2;
}
else
{
left=mid+1;
mid=(left+right)/2;
}
}

}
}
return false;
}
};

第二种思路：从右上角往左下角搜索，O(m+n)

class Solution {//O(Mlog(N)
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
int m=matrix.size();
if(m==0) return false;
int n=matrix[0].size();
int i=0,j=n-1;
while(i<m&&j>=0)
{
if(matrix[i][j]>target)
j--;
else if(matrix[i][j]<target)
i++;
else
return true;

}
return false;
}
};