题目：

给定一个正整数 n，生成一个包含 1 到 n2 所有元素，且元素按顺时针顺序螺旋排列的正方形矩阵。

示例:

输入：3

输出：

[ [1,2,3],

  [4,5,6],

  [7,8,9]

]

 

解题思路：

参见leetcode54

 

代码1：

class Solution {public:    vector
     
      
       > generateMatrix(int n) {        vector
       
         v(n,0);        vector
        
         
          > res(n,v);        vector
          
           
            > res_flag(n,v); int i=0; int j=0; int have_left = 0; int have_right = 0; int have_up = 0; int have_down = 0; string cur_direction = "right"; for(int num=1;num<=n*n;num++){ res[i][j] = num; res_flag[i][j] = 1; if (j - 1 >= 0 && res_flag[i][j - 1] == 0) have_left = 1; if (j + 1 < n && res_flag[i][j + 1] == 0) have_right = 1; if (i - 1 >= 0 && res_flag[i - 1][j] == 0) have_up = 1; if (i + 1 < n && res_flag[i + 1][j] == 0) have_down = 1; cur_direction = nextDirection(cur_direction, have_left,have_right,have_up,have_down); if(cur_direction == "left") j--; if(cur_direction == "right") j++; if(cur_direction == "up") i--; if(cur_direction == "down") i++; have_left = 0; have_right = 0; have_up = 0; have_down = 0; } return res; } string nextDirection(string cur_direction,int have_left,int have_right,int have_up,int have_down){ if(cur_direction == "left" && have_left == 1) return "left"; if(cur_direction == "right" && have_right == 1) return "right"; if(cur_direction == "up" && have_up == 1) return "up"; if(cur_direction == "down" && have_down == 1) return "down"; if(have_left == 1) return "left"; if(have_right == 1) return "right"; if(have_up == 1) return "up"; if(have_down == 1) return "down"; return "nonext"; }};
           
          
         
        
       
      
     

 

代码2：

class Solution {public:    vector
     
      
       > generateMatrix(int n) {        vector
       
        
         > res(n, vector
         
          (n, 0));        int up = 0, down = n - 1, left = 0, right = n - 1, val = 1;        while (true) {            for (int j = left; j <= right; ++j) res[up][j] = val++;            if (++up > down) break;            for (int i = up; i <= down; ++i) res[i][right] = val++;            if (--right < left) break;            for (int j = right; j >= left; --j) res[down][j] = val++;            if (--down < up) break;            for (int i = down; i >= up; --i) res[i][left] = val++;            if (++left > right) break;        }        return res;    }};