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Spiral Matrix I II

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].

题解:

class Solution {public:    vector<int> spiralOrder(vector<vector<int> > &matrix) {        vector<int> res;        int row = matrix.size();        if(row==0)            return res;        int col = matrix[0].size();        int k = (1+min(row, col))/2;        int row0 = 0, col0 = 0;        int i=0, j=0;        while(k--) {            if((row-row0)==1)                for(j=col0;j<col;j++)                    res.push_back(matrix[row0][j]);            else if((col-col0)==1)                for(i=row0;i<row;i++)                    res.push_back(matrix[i][col-1]);            else {                for(j=col0;j<col;j++)                     res.push_back(matrix[row0][j]);                res.pop_back();                for(i=row0;i<row;i++)                     res.push_back(matrix[i][col-1]);                res.pop_back();                for(j=col-1;j>=col0;j--)                     res.push_back(matrix[row-1][j]);                res.pop_back();                for(i=row-1;i>=row0;i--)                     res.push_back(matrix[i][col0]);                res.pop_back();                row0++;col0++;                row--;col--;            }        }        return res;    }};

 

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 ]]
题解:
class Solution {public:    vector<vector<int> > generateMatrix(int n) {        vector<vector<int> > res(n, vector<int>(n));        int start = 0;        int end = n-1;        int i=0, j=0;        int num = 1;        while(start<end) {            for(j=start;j<end;j++)                res[start][j] = num++;            for(i=start;i<end;i++)                res[i][end] = num++;            for(j=end;j>start;j--)                res[end][j] = num++;            for(i=end;i>start;i--)                res[i][start] = num++;            start++;end--;        }        if(start==end)            res[end][end] = num;        return res;    }};

 

Spiral Matrix I II