LeetCode 54/59. Spiral Matrix i, ii

1. 题目描述

54

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].

59

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 ]

]

2. 解题思路

这两道题目非常类似， 处理的思路也是基本雷同， 只需要处理下访问一周的数据方法， 即可

3. code

3.1 54

class Solution {
public:
vector<int> spiralOrder(vector<vector<int>>& matrix) {
vector<int> res;
if (matrix.size() == 0)
return res;

int m = matrix.size();
int n = matrix[0].size();
for (int i = 0; i != (min(m, n) + 1) / 2; i++)
search_around(matrix, m, n, i, res);

return res;
}

private:
void search_around(vector<vector<int>> & arr, int m, int n, int offset, vector<int> & res){
if (offset == m - offset - 1 && m <= n){
for (int i = offset; i != n - offset; i++)
res.push_back(arr[offset][i]);
return;
}

if (offset == n - offset - 1 && m >= n){
for (int i = offset; i != m - offset; i++)
res.push_back(arr[i][offset]);
return;
}

for (int j = offset; j != n - offset - 1; j++)
res.push_back(arr[offset][j]);

for (int i = offset; i != m - offset - 1; i++)
res.push_back(arr[i][n - offset - 1]);

for (int j = n - offset - 1; j != offset; j--)
res.push_back(arr[m - offset - 1][j]);

for (int i = m - offset - 1; i != offset; i--)
res.push_back(arr[i][offset]);
}
};

3.2 59

class Solution {
public:
vector<vector<int>> generateMatrix(int n) {
vector<vector<int>> arr(n, vector<int>(n, 0));
int cur_depth = 1;
for (int i = 0; i != (n + 1) / 2; i++)
search_around(arr, n, cur_depth, i);

return arr;
}

private:
void search_around(vector<vector<int>> & arr, int n, int & cur_depth, int offset){
if (offset == n - offset - 1){
arr[offset][offset] = cur_depth++;
return;
}

for (int j = offset; j != n - offset - 1; j++)
arr[offset][j] = cur_depth++;

for (int i = offset; i != n - offset - 1; i++)
arr[i][n - offset - 1] = cur_depth++;

for (int j = n - offset - 1; j != offset; j--)
arr[n - offset - 1][j] = cur_depth++;

for (int i = n - offset - 1; i != offset; i--)
arr[i][offset] = cur_depth++;
}
};

4. 大神解法

4.2 54

一样的思想， 不过感觉个人将 访问一周的数据独立成一个函数更加容易理解一些

/*
This is a very simple and easy to understand solution. I traverse right and increment rowBegin, then traverse down and decrement colEnd, then I traverse left and decrement rowEnd, and finally I traverse up and increment colBegin.

The only tricky part is that when I traverse left or up I have to check whether the row or col still exists to prevent duplicates. If anyone can do the same thing without that check, please let me know!

Any comments greatly appreciated.
*/

public class Solution {
public List<Integer> spiralOrder(int[][] matrix) {

List<Integer> res = new ArrayList<Integer>();

if (matrix.length == 0) {
return res;
}

int rowBegin = 0;
int rowEnd = matrix.length-1;
int colBegin = 0;
int colEnd = matrix[0].length - 1;

while (rowBegin <= rowEnd && colBegin <= colEnd) {
// Traverse Right
for (int j = colBegin; j <= colEnd; j ++) {
res.add(matrix[rowBegin][j]);
}
rowBegin++;

// Traverse Down
for (int j = rowBegin; j <= rowEnd; j ++) {
res.add(matrix[j][colEnd]);
}
colEnd--;

if (rowBegin <= rowEnd) {
// Traverse Left
for (int j = colEnd; j >= colBegin; j --) {
res.add(matrix[rowEnd][j]);
}
}
rowEnd--;

if (colBegin <= colEnd) {
// Traver Up
for (int j = rowEnd; j >= rowBegin; j --) {
res.add(matrix[j][colBegin]);
}
}
colBegin ++;
}

return res;
}
}

4.2 59

clean logic

class Solution {
public:
vector<vector<int> > generateMatrix(int n) {
vector<vector<int> > ret( n, vector<int>(n) );
int k = 1, i = 0;
while( k <= n * n )
{
int j = i;
// four steps
while( j < n - i )             // 1. horizonal, left to right
ret[i][j++] = k++;
j = i + 1;
while( j < n - i )             // 2. vertical, top to bottom
ret[j++][n-i-1] = k++;
j = n - i - 2;
while( j > i )                  // 3. horizonal, right to left
ret[n-i-1][j--] = k++;
j = n - i - 1;
while( j > i )                  // 4. vertical, bottom to  top
ret[j--][i] = k++;
i++;      // next loop
}
return ret;
}
};