LeetCode | 59. 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 ]]

题意：填充蛇形矩阵

class Solution {public:/*    蛇形矩阵*/    vector<vector<int>> generateMatrix(int n)    {        vector<vector<int> > matrix;        int temp = 0;        for(int i=0;i<n;i++)        {            vector<int> line;            for(int j=0;j<n;j++)            {                line.push_back(temp);            }            matrix.push_back(line);        }        vector<int> res;        int x = 0, y = 0, tot = 1;  //tot用于计数，作为停止条件        int up = 0, down = n-1, left = 0, right = n-1;        if(n==0)        //错误输入数据，特殊处理            return matrix;        //开始遍历        while(tot <= n*n)        {            while(x==up && tot <= n*n)     //最上面一行            {                matrix[x][y] = tot;                tot++;                if(y == right)                    x++;                else                    y++;            }            up += 1;            while(y==right  && tot <= n*n)            {                matrix[x][y] = tot;                tot++;                if(x == down)                    y--;                else                    x++;            }            right -= 1;            while(x==down && tot <= n*n)            {                matrix[x][y] = tot;                tot++;                if(y == left)                    x--;                else                    y--;            }            down -= 1;            while(y==left && tot <= n*n)            {                matrix[x][y] = tot;                tot++;                if(x == up)                    y++;                else                    x--;            }            left += 1;        }        //cout<<up<<down<<left<<right;        return matrix;    }};

附上solution区代码：

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;        }    };