【LeetCode】Search a 2D Matrix & Set Matrix Zeroes & Spiral Matrix & Spiral Matrix II

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

.

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

Set Matrix Zeroes

 

Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place.
click to show follow up.
Follow up:
Did you use extra space?

A straight forward solution using O(mn) space is probably a bad idea.

A simple improvement uses O(m + n) space, but still not the best solution.

Could you devise a constant space solution?

class Solution {
public:
void setZeroes(vector<vector<int> > &matrix) {
if(matrix.size() < 1)	return;
vector<int > col(matrix[0].size(),0);
vector<int > row(matrix.size());
for(int i = 0;i<matrix.size();i++)
{
for(int j = 0;j<matrix[0].size();j++)
{
if(matrix[i][j] == 0)
{
row[i] = 1;
col[j] = 1;
}
}
}
for(int i = 0;i<matrix.size();i++)
for(int j = 0;j<matrix[0].size();j++)
if(row[i] == 1 || col[j] == 1)
matrix[i][j] = 0;
}
};

Spiral Matrix

 

Given a matrix of m x n elements (m rows, n columns), return all elements of
the matrix in spiral order.
For example,

Given the following matrix:

[
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]
]

You should return 

[1,2,3,6,9,8,7,4,5]

.

class Solution {
public:
vector<int> spiralOrder(vector<vector<int> > &matrix) {
vector<int> ret;
if(matrix.size()<1)	return ret;
for(int i = 0;i<(min(matrix.size(),matrix[0].size())+1)/2;i++)
spiralOrder(ret,matrix,i);
return ret;
}
void spiralOrder(vector<int> &ret,vector<vector<int> > &matrix,int start)
{//左上到右上
int m =  matrix.size(), n = matrix[0].size();
for(int i = start;i<n-start;i++)
ret.push_back(matrix[start][i]);
if(m - start*2 == 1) return;/////////////////！！！！！！
//右上到右下
for(int i = start+1;i<m-start;i++)
ret.push_back(matrix[i][n-start-1]);
if(n - start*2 == 1) return;////////////////////！！！！！！
//右下到左下
for(int i = n-start-2;i>= start;i--)
ret.push_back(matrix[m-start-1][i]);
//右下到左上
for(int i = m-start-2;i> start;i--)
ret.push_back(matrix[i][start]);
}
};

Spiral Matrix II

 

Given an integer n, generate a square matrix filled with elements from 1 to n2 in
spiral order.
For example,

Given n = 

3

,
You should return the following matrix:

[
[ 1, 2, 3 ],
[ 8, 9, 4 ],
[ 7, 6, 5 ]
]

class Solution {
public:
vector<vector<int> > generateMatrix(int n) {
vector<vector<int> > ret(n,vector<int> (n));
int sum = 1;
for(int i = 0;i<(n+1)/2;i++)
generateMatrixCore(ret,sum,i);
return ret;
}
void generateMatrixCore(vector<vector<int> > &ret,int &sum,int start)
{//左上到右上
int n = ret.size();
for(int i = start;i<n-start;i++)
ret[start][i] = sum++;
if(n - start*2 == 1) return;
//右上到右下
for(int i = start+1;i<n-start;i++)
ret[i][n-start-1] = sum++;
//右下到左下
for(int i = n-start-2;i>= start;i--)
ret[n-start-1][i] = sum++;
//右下到左上
for(int i = n-start-2;i> start;i--)
ret[i][start] = sum++;
}
};

推荐学习C++的资料

C++标准函数库
http://download.csdn.net/detail/chinasnowwolf/7108919

在线C++API查询
http://www.cplusplus.com/

vector使用方法

http://www.cplusplus.com/reference/vector/vector/