spiral-matrix-ii &i 生成顺时针序列

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<int> spiralOrder(vector<vector<int> > &matrix) {
        int m=matrix.size();
        vector<int> 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<vector<int> > generateMatrix(int n) {
        vector<vector<int>> result(n,vector<int>(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;
    }
};