[LeetCode] 54. 螺旋矩阵

题目链接 : https://leetcode-cn.com/problems/spiral-matrix/

题目描述:

给定一个包含 m x n 个元素的矩阵（m 行, n 列），请按照顺时针螺旋顺序，返回矩阵中的所有元素。

示例:

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

示例 2:

输入:
[
  [1, 2, 3, 4],
  [5, 6, 7, 8],
  [9,10,11,12]
]
输出: [1,2,3,4,8,12,11,10,9,5,6,7]

思路:

思路一:模拟过程

通过控制行的上下边界,列的左右边界

思路二: 旋转

直接举例子,

1  2  3   弹出第一行,逆时针旋转    6  9 重复操作  8  7
4  5  6   ------------------>   5  8 ---->   5  4 --->...
7  8  9                         4  7

把弹出输出即可.

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代码:

思路一:

python

class Solution:
    def spiralOrder(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: List[int]
        """
        if not matrix : return []
        shang_row = 0
        xia_row = len(matrix) - 1 
        zuo_col = 0
        you_col = len(matrix[0]) - 1
        res = []
        while shang_row <= xia_row  and  zuo_col <= you_col:
            # 从左到右
            for i in range(zuo_col, you_col+1):
                #print(shang_row, i)
                res.append(matrix[shang_row][i])
            shang_row += 1
            if shang_row > xia_row:break
            # 从上到下
            for i in range(shang_row, xia_row+1):
                res.append(matrix[i][you_col])
            you_col -= 1
            if zuo_col > you_col:break
            # 从右到左
            for i in range(you_col, zuo_col - 1,-1):
                #print(xia_row - 1, i)
                res.append(matrix[xia_row][i])
            xia_row -= 1
            # 从下到上
            for i in range(xia_row , shang_row - 1, -1):
                #print(i, zuo_col)  
                res.append(matrix[i][zuo_col])
            zuo_col += 1
        return res

java

class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        if (matrix == null || matrix.length == 0) return new ArrayList<>();
        List<Integer> res = new ArrayList<>();
        int shang_row = 0;
        int xia_row = matrix.length - 1;
        int zou_col = 0;
        int you_col = matrix[0].length - 1;
        while (shang_row <= xia_row && zou_col <= you_col) {
            // 从左到右
            for (int i = zou_col; i <= you_col; i++) res.add(matrix[shang_row][i]);
            shang_row++;
            if (shang_row > xia_row) break;
            // 从上到下
            for (int i = shang_row; i <= xia_row; i++) res.add(matrix[i][you_col]);
            you_col--;
            if (zou_col > you_col) break;
            // 从右到左
            for (int i = you_col; i >= zou_col; i--) res.add(matrix[xia_row][i]);
            xia_row--;
            //从下到上
            for (int i = xia_row; i >= shang_row; i--) res.add(matrix[i][zou_col]);
            zou_col++;
        }
        return res;
    }
}

思路二:

class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        if not matrix : return []
        res = []
        while matrix:
            res.extend(matrix.pop(0))
            next_matrix = []
            #print(matrix)
            for x in zip(*matrix):
                next_matrix.append(x)
            #print(next_matrix)
            matrix = next_matrix[::-1]
        return res