[LeetCode] 54. Spiral Matrix 螺旋矩阵

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].

将一个矩阵按螺旋顺序输出。

解法：用变量left, right, top, bottom记录左，右，顶，底。然后按照左到右，顶到底，右到左，底到顶的顺序循环，把遍历的元素加入到结果。

Java:

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public class Solution {     public List<Integer> spiralOrder(int[][] matrix) {         List<Integer> res = new ArrayList<>();         if(matrix == null || matrix.length == 0)             return res;         int rowNum = matrix.length, colNum = matrix[0].length;         int left = 0, right = colNum - 1, top = 0, bot = rowNum - 1;                   while(res.size() < rowNum * colNum) {             for(int col = left; col <= right; col++)                 res.add(matrix[top][col]);             top++;             if(res.size() < rowNum * colNum) {                 for(int row = top; row <= bot; row++)                     res.add(matrix[row][right]);                 right--;                }             if(res.size() < rowNum * colNum) {                 for(int col = right; col >= left; col--)                     res.add(matrix[bot][col]);                 bot--;             }             if(res.size() < rowNum * colNum) {                 for(int row = bot; row >= top; row--)                     res.add(matrix[row][left]);                 left++;             }         }                   return res;     } }

Python:

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class Solution:     # @param matrix, a list of lists of integers     # @return a list of integers     def spiralOrder(self, matrix):         result = []         if matrix == []:             return result                   left, right, top, bottom = 0, len(matrix[0]) - 1, 0, len(matrix) - 1                   while left <= right and top <= bottom:             for j in xrange(left, right + 1):                 result.append(matrix[top][j])             for i in xrange(top + 1, bottom):                 result.append(matrix[i][right])             for j in reversed(xrange(left, right + 1)):                 if top < bottom:                     result.append(matrix[bottom][j])             for i in reversed(xrange(top + 1, bottom)):                 if left < right:                     result.append(matrix[i][left])             left, right, top, bottom = left + 1, right - 1, top + 1, bottom - 1                       return result

C++：

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class Solution { public:     vector<int> spiralOrder(vector<vector<int> > &matrix) {         vector<int> res;         if (matrix.empty() || matrix[0].empty()) return res;         int m = matrix.size(), n = matrix[0].size();         int c = m > n ? (n + 1) / 2 : (m + 1) / 2;         int p = m, q = n;         for (int i = 0; i < c; ++i, p -= 2, q -= 2) {             for (int col = i; col < i + q; ++col)                 res.push_back(matrix[i][col]);             for (int row = i + 1; row < i + p; ++row)                 res.push_back(matrix[row][i + q - 1]);             if (p == 1 || q == 1) break;             for (int col = i + q - 2; col >= i; --col)                 res.push_back(matrix[i + p - 1][col]);             for (int row = i + p - 2; row > i; --row)                 res.push_back(matrix[row][i]);         }         return res;     } };

　　

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