【Spiral Matrix II】cpp

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 ]
]

代码：

class Solution {
public:
    vector<vector<int>> generateMatrix(int n) {
            vector<vector<int> > ret(n, vector<int>(n,0));
            int circle = n/2;
            int v = 1;
            for ( int c=0; c<circle; ++c )
            {
                // up
                for ( int col=c; col<n-c; ++col ) ret[c][col]=v++;
                // right
                for ( int row=c+1; row<n-c-1; ++row ) ret[row][n-1-c]=v++;
                // down
                for ( int col=n-1-c; col>=c; --col ) ret[n-1-c][col]=v++;
                // left
                for ( int row=n-2-c; row>c; --row) ret[row][c]=v++;
            }
            if ( n & 1 )
            {
                ret[circle][circle]=v;
            }
            return ret;
    }
};

tips：

按照Spiral Matrix的顺序走一遍元素，维护一个v每次运算后+1。

============================================

第二次过这道题，思路跟第一次一样。

class Solution {
public:
    vector<vector<int>> generateMatrix(int n) {
            vector<vector<int> > ret(n, vector<int>(n,1));
            if ( n<1 ) return ret;
            int val = 1;
            for ( int i=0; i<n/2; ++i )
            {
                // north
                for ( int p=i; p<n-i; ++p ) ret[i][p] = val++; 
                // east
                for ( int p=i+1; p<n-1-i; ++p ) ret[p][n-1-i] = val++;
                // south
                for ( int p=i; p<n-i; ++p ) ret[n-1-i][n-1-p] = val++;
                // west
                for ( int p=i+1; p<n-1-i; ++p ) ret[n-1-p][i] = val++;
            }
            if ( n & 1 ) ret[n/2][n/2] = val;
            return ret;
    }
};