Spiral Matrix II

题目：

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 ]
]

CPP:

class Solution {
public:
    vector<vector<int>> generateMatrix(int x) {
        vector<vector<int>> result;
        if (x == 0) {
            return result;
        } else {
            vector<int> v(x);
            for (int i = 0; i < x; i++) {
                result.push_back(v);
            }
        }

        int r = 0; //第几圈
        int i;
        //left to right
        int a = 1;
        while (r < x / 2) {
            for (i = r; i < x - r; i++) {
                result[r][i] = a;
                a++;
            }
            //up to down
            for (i = r + 1; i < x - r ; i++) {
                result[i][x - r-1] = a;
                a++;
            }
            //right to left
            for (i = x - r - 2; i >= r; i--) {
                result[x - r-1][i] = a;
                a++;
            }
            //down to up
            for (i = x - r - 2; i > r; i--) {
                result[i][r] = a;
                a++;
            }
            r++;
        }
        if(2*r !=x)
        {
            result[r][r] = a;
        }
        return result;
    }
};

　python:

 

class Solution(object):
    def generateMatrix(self,n):
        ans = []
        for x in range(n):
            ans.append([0]*n)
        mini,minj,maxi,maxj = 0,0,n-1,n-1
        num = 1
        
        while maxi >= mini and maxj >= minj:
            #left to right
            for j in range(minj,maxj+1):
                ans[minj][j] = num
                num +=1
            mini += 1
            #up to down
            for i in range(mini,maxi+1):
                ans[i][maxj] = num
                num +=1
            maxj -= 1
            #right to left 
            for j in range(maxj,minj-1,-1):
                ans[maxi][j] = num
                num +=1
            maxi -=1
            #down to up
            for i in range(maxi,mini-1,-1):
                ans[i][minj] = num
                num +=1
            minj += 1
        return ans;