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【leetcode_medium】54. Spiral Matrix

leetcode_medium_array

problem

​​54. Spiral Matrix​​

solution #1: traverse from left to right, and then from down to up;

code

 

class Solution {
public:
vector<int> spiralOrder(vector<vector<int>>& matrix) {
vector<int> res;
if(matrix.empty()) return res;
int m = matrix.size();
int n = matrix[0].size();
int up = 0, down = m-1, left = 0, right = n-1;
while(left<=right && up<=down ) // && res.size()<m*n
{
// why do check res.size in every for loop?
for(int i=left; i<=right&& res.size()<m*n; i++)
{
res.push_back(matrix[up][i]);
}
up++;
for(int i=up; i<=down&& res.size()<m*n; i++)
{
res.push_back(matrix[i][right]);
}
right--;
for(int i=right; i>=left&& res.size()<m*n; i--)
{
res.push_back(matrix[down][i]);
}
down--;
for(int i=down; i>=up&& res.size()<m*n; i--)
{
res.push_back(matrix[i][left]);
}
left++;
}
return res;
}
};

  

class Solution {
public:
vector<int> spiralOrder(vector<vector<int>>& matrix) {
vector<int> res;
if(matrix.empty()) return res;
int m = matrix.size();
int n = matrix[0].size();
int up = 0, down = m-1, left = 0, right = n-1;
while(left<=right && up<=down ) // && res.size()<m*n
{
// why do check res.size in every for loop?
for(int i=left; i<=right&& res.size()<m*n; i++)
{
res.push_back(matrix[up][i]);
}
up++;
for(int i=up; i<=down&& res.size()<m*n; i++)
{
res.push_back(matrix[i][right]);
}
right--;
for(int i=right; i>=left&& res.size()<m*n; i--)
{
res.push_back(matrix[down][i]);
}
down--;
for(int i=down; i>=up&& res.size()<m*n; i--)
{
res.push_back(matrix[i][left]);
}
left++;
}
return res;
}
};

 

solution #2:

code 

 

solution #3:使用switch语句;

code:

 

参考

1. ​​leetcode_54. Spiral Matrix​​;

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