I:

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

For example,

Given the following matrix:

[ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ]]

You should return[1,2,3,6,9,8,7,4,5].

按顺时针取序列，因为序列不一定是正矩阵，所以需要每取完一个方向要当即--或++，并做判断是否需要再取下一个方向。

当取完left->right,需要top++表明上一行已取完，top->bottom需right--，right->left需先判断top<=bottom避免只有单行重复取了上行，而后bottom++，bottom->top需先判断left<=right避免重复取右列，而后left++;

class Solution {public:    vector
     
       spiralOrder(vector
      
       
         > &matrix) {        int m=matrix.size();        vector
        
          res;        if(m==0)            return res;        int n=matrix[0].size();        int left=0,right=n-1,top=0,bottom=m-1;        int i=0;        while(left<=right&&top<=bottom)        {            for(i=left;i<=right;++i)                res.push_back(matrix[top][i]);            top++;            for(i=top;i<=bottom;++i)                res.push_back(matrix[i][right]);            right--;            if(top<=bottom){            for(i=right;i>=left;--i)                res.push_back(matrix[bottom][i]);            }            bottom--;            if(left<=right){            for(i=bottom;i>=top;--i)                res.push_back(matrix[i][left]);            }            left++;        }        return res;    }};
        
       
      
     

 

II:

Given an integer n, generate a square matrix filled with elements from 1 to n 2 in spiral order.

For example,

Given n =3,

You should return the following matrix:

[ [ 1, 2, 3 ], [ 8, 9, 4 ], [ 7, 6, 5 ]] 注意左->右可遍历全，剩余两个方向遍历时需+1，最后一个方向掐头去尾避免重复遍历

class Solution {public:    vector
     
      
        > generateMatrix(int n) {        vector
       
        
         > result(n,vector
         
          (n));                 if(n==0)            return result;                 int step = 1;                 int left = 0; int right = n-1; int top = 0; int bottom = n-1;        while(left<=right && top<=bottom)        {            // 左->右            for(int i=left;i<=right;i++) {                result[top][i] = step;                step++;            }            //上->下            for(int i=top+1;i<=bottom;i++) {                result[i][right] = step;                step++;            }            //右->左            for(int i=right-1;i>=left;i--) {                result[bottom][i] = step;                step++;            }            //下->上            for(int i=bottom-1;i>top;i--) {                result[i][left] = step;                step++;            }                         left++; right--; top++; bottom--;        }        return result;    }};