6、螺旋矩阵

题目描述

54. 螺旋矩阵

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。

示例一

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

示例二

输入：matrix = [[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]

题目解析

这个题目很细非常细，思路很容易想到，但是要是完全实现也不是特别容易，我们一起分析下这个题目，我们可以这样理解，我们像剥洋葱似的一步步的剥掉外皮，直到遍历结束，见下图。

*

题目很容易理解，但是要想完全执行出来，也是不容易的，因为这里面的细节太多了，我们需要认真仔细的考虑边界。

我们也要考虑重复遍历的情况即什么时候跳出循环。刚才我们通过箭头知道了我们元素的遍历顺序，这个题目也就完成了一大半了，下面我们来讨论一下什么时候跳出循环，见下图。

注：这里需要注意的是，框框代表的是每个边界。

题目代码

Java Code:

class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {

        List<Integer> arr = new ArrayList<>();
        int left = 0, right = matrix[0].length-1;
        int top = 0, down = matrix.length-1;

        while (true) {
             for (int i = left; i <= right; ++i) {
                 arr.add(matrix[top][i]);
             }
             top++;
             if (top > down) break;
             for (int i = top; i <= down; ++i) {
                 arr.add(matrix[i][right]);
             }
             right--;
             if (left > right) break;
             for (int i = right; i >= left; --i) {
                 arr.add(matrix[down][i]);
             }
             down--;
             if (top > down) break;
             for (int i = down; i >= top; --i) {
                 arr.add(matrix[i][left]);
             }
             left++;
             if (left > right) break;

        }
        return arr;
    }
}

C++ Code:

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        vector <int> arr;
        int left = 0, right = matrix[0].size()-1;
        int top = 0, down = matrix.size()-1;
        while (true) {
             for (int i = left; i <= right; ++i) {
                 arr.emplace_back(matrix[top][i]);
             }
             top++;
             if (top > down) break;
             for (int i = top; i <= down; ++i) {
                 arr.emplace_back(matrix[i][right]);
             }
             right--;
             if (left > right) break;
             for (int i = right; i >= left; --i) {
                 arr.emplace_back(matrix[down][i]);
             }
             down--;
             if (top > down) break;
             for (int i = down; i >= top; --i) {
                 arr.emplace_back(matrix[i][left]);
             }
             left++;
             if (left > right) break;
        }
        return arr;
    }
};

Python3 Code:

from typing import List
class Solution:
    def spiralOrder(self, matrix: List[List[int]])->List[int]:
        arr = []
        left = 0
        right = len(matrix[0]) - 1
        top = 0
        down = len(matrix) - 1
        while True:
            for i in range(left, right + 1):
                arr.append(matrix[top][i])
            top += 1
            if top > down:
                break
            for i in range(top, down + 1):
                arr.append(matrix[i][right])
            right -= 1
            if left > right:
                break
            for i in range(right, left - 1, -1):
                arr.append(matrix[down][i])
            down -= 1
            if top > down:
                break
            for i in range(down, top - 1, -1):
                arr.append(matrix[i][left])
            left += 1
            if left > right:
                break
        return arr

Swift Code

class Solution {
    func spiralOrder(_ matrix: [[Int]]) -> [Int] {
        var arr:[Int] = []
        var left = 0, right = matrix[0].count - 1
        var top = 0, down = matrix.count - 1

        while (true) {
            for i in left...right {
                arr.append(matrix[top][i])
            }
            top += 1
            if top > down { break }
            for i in top...down {
                arr.append(matrix[i][right])
            }
            right -= 1
            if left > right { break}
            for i in stride(from: right, through: left, by: -1) {
                arr.append(matrix[down][i])
            }
            down -= 1
            if top > down { break}
            for i in stride(from: down, through: top, by: -1) {
                arr.append(matrix[i][left])
            }
            left += 1
            if left > right { break}
        }

        return arr
    }
}

Go Code:

func spiralOrder(matrix [][]int) []int {
    res := []int{}
    left, right := 0, len(matrix[0]) - 1
    top, down   := 0, len(matrix) - 1

    for {
        for i := left; i <= right; i++ {
            res = append(res, matrix[top][i])
        }
        top++
        if top > down { break }

        for i := top; i <= down; i++ {
            res = append(res, matrix[i][right])
        }
        right--
        if left > right { break }

        for i := right; i >= left; i-- {
            res = append(res, matrix[down][i])
        }
        down--
        if top > down { break }

        for i := down; i >= top; i-- {
            res = append(res, matrix[i][left])
        }
        left++
        if left > right { break }
    }
    return res
}