【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,));

            int circle = n/;

            int v = ;

            for ( int c=; c<circle; ++c )

            {

                // up

                for ( int col=c; col<n-c; ++col ) ret[c][col]=v++;

                // right

                for ( int row=c+; row<n-c-; ++row ) ret[row][n--c]=v++;

                // down

                for ( int col=n--c; col>=c; --col ) ret[n--c][col]=v++;

                // left

                for ( int row=n--c; row>c; --row) ret[row][c]=v++;

            }

            if ( n &  )

            {

                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,));

            if ( n< ) return ret;

            int val = ;

            for ( int i=; i<n/; ++i )

            {

                // north

                for ( int p=i; p<n-i; ++p ) ret[i][p] = val++;

                // east

                for ( int p=i+; p<n--i; ++p ) ret[p][n--i] = val++;

                // south

                for ( int p=i; p<n-i; ++p ) ret[n--i][n--p] = val++;

                // west

                for ( int p=i+; p<n--i; ++p ) ret[n--p][i] = val++;

            }

            if ( n &  ) ret[n/][n/] = val;

            return ret;

    }

};

巴特西

【Spiral Matrix II】cpp

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