LeetCode:Spiral Matrix I II

Spiral Matrix

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].

打印螺旋矩阵

逐个环的打印, 对于m *n的矩阵,环的个数是 (min(n,m)+1) / 2。对于每个环顺时针打印四条边。

注意的是：最后一个环可能只包含一行或者一列数据

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int> > &matrix) {
        int m = matrix.size(), n;
        if(m != 0)n = matrix[0].size();
        int cycle = m > n ? (n+1)/2 : (m+1)/2;//环的数目
        vector<int>res;
        
        int a = n, b = m;//a,b分别为当前环的宽度、高度
        for(int i = 0; i < cycle; i++, a -= 2, b -= 2)
        {
            //每个环的左上角起点是matrix[i][i],下面顺时针依次打印环的四条边
            for(int column = i; column < i+a; column++)
                res.push_back(matrix[i][column]);
            for(int row = i+1; row < i+b; row++)
                res.push_back(matrix[row][i+a-1]);
            if(a == 1 || b == 1)break; //最后一个环只有一行或者一列
            for(int column = i+a-2; column >= i; column--)
                res.push_back(matrix[i+b-1][column]);
            for(int row = i+b-2; row > i; row--)
                res.push_back(matrix[row][i]);
        }
        return res;
    }
};

---

Spiral Matrix II

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 ]
]

本质上和上一题是一样的，这里我们要用数字螺旋的去填充矩阵。同理，我们也是逐个环的填充，每个环顺时针逐条边填充                 本文地址

class Solution {
public:
    vector<vector<int> > generateMatrix(int n) {
        vector<vector<int> > matrix(n, vector<int>(n));
        int a = n;//a为当前环的边长
        int val = 1;
        for(int i = 0; i < n/2; i++, a -= 2)
        {
            //每个环的左上角起点是matrix[i][i],下面顺时针依次填充环的四条边
            for(int column = i; column < i+a; column++)
                matrix[i][column] = val++;
            for(int row = i+1; row < i+a; row++)
                matrix[row][i+a-1] = val++;
            for(int column = i+a-2; column >= i; column--)
                matrix[i+a-1][column] = val++;
            for(int row = i+a-2; row > i; row--)
                matrix[row][i] = val++;
        }
        if(n % 2)matrix[n/2][n/2] = val;//n是奇数时，最后一个环只有一个数字
        return matrix;
    }
};

【版权声明】转载请注明出处：http://www.cnblogs.com/TenosDoIt/p/3774747.html