LeetCode-54、59：Spiral Matrix (螺旋输出矩阵)

题目54：Spiral Matrix

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

例子：

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].

问题解析：

给定m*n大小的矩阵，螺旋输出矩阵的元素。

链接：

• https://leetcode.com/problems/spiral-matrix/description/

思路标签：

数据结构：Vector

螺旋输出就OK

解答：

• 每个旋涡圈都有上、下、左、右四个边界线；
• 每输出一个边界，相应的边界线值+1或者-1；
• 直接旋涡输出就OK；
• 注意边界的判定。

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        if(matrix.size() <= 0) return {};
        int rows = matrix.size();
        int cols = matrix[0].size();
        vector<int> spiral(rows*cols);
        int u = 0, d = rows-1, l = 0, r = cols-1, k = 0;
        while(true){
            for(int col = l; col <= r; ++col)
                spiral[k++] = matrix[u][col];
            if(++u > d) break;

            for(int row = u; row <= d; ++row)
                spiral[k++] = matrix[row][r];
            if(--r < l) break;

            for(int col = r; col >= l; --col)
                spiral[k++] = matrix[d][col];
            if(--d < u) break;

            for(int row = d; row >= u; --row)
                spiral[k++] = matrix[row][l];
            if(++l > r) break;
        }

        return spiral;
    }
};

---

题目59：Unique Paths II

Given an integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.

例子：

For example,
Given n = 3,

You should return the following matrix:

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

问题解析：

给定数字n，构造n*n的螺旋矩阵。

链接：

• https://leetcode.com/problems/spiral-matrix-ii/description/

思路标签：

数据结构：Vector

解答：

• 与54的上下左右边界的做法是相似的。

class Solution {
public:
    vector<vector<int>> generateMatrix(int n) {
        if(n < 1) return {};
        vector<vector<int> > spiral(n, vector<int>(n));
        int u = 0, d = n-1, l = 0, r = n-1;
        int k = 1;
        while(k <= n*n){
            for(int col = l; col <= r; ++col)
                spiral[u][col] = k++;
            u++;

            for(int row = u; row <= d; ++row)
                spiral[row][r] = k++;
            r--;

            for(int col = r; col >= l; --col)
                spiral[d][col] = k++;
            d--;

            for(int row = d; row >= u; --row)
                spiral[row][l] = k++;
            l++;
        }

        return spiral;
    }
};