algorithm-base/animation-simulation/数组篇/leetcode59螺旋矩阵2.md

如果阅读时，发现错误，或者动画不可以显示的问题可以添加我微信好友 tan45du_one ，备注 github + 题目 + 问题 向我反馈

感谢支持，该仓库会一直维护，希望对各位有一丢丢帮助。

另外希望手机阅读的同学可以来我的 公众号：程序厨 两个平台同步，想要和题友一起刷题，互相监督的同学，可以在我的小屋点击刷题小队进入。

59.螺旋矩阵 II

给你一个正整数 n ，生成一个包含 1 到 n2 所有元素，且元素按顺时针顺序螺旋排列的 n x n 正方形矩阵 matrix 。

示例 1：

输入：n = 3 输出：1,2,3],[8,9,4],[7,6,5

示例 2：

输入：n = 1 输出：1

其实我们只要做过了螺旋矩阵 第一题，这个题目我们完全可以一下搞定，几乎没有进行更改，我们先来看下 leetcode 54 题的解析。

leetcode 54 螺旋矩阵

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]

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;
    }
}

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

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;
    }
};

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
    }
}

我们仅仅是将 54 反过来了，往螺旋矩阵里面插值，下面我们直接看代码吧,大家可以也可以对其改进，大家可以思考一下，如果修改能够让代码更简洁！

Java Code:

class Solution {
    public int[][] generateMatrix(int n) {

        int[][] arr = new int[n][n];
        int left = 0;
        int right = n-1;
        int top = 0;
        int buttom = n-1;
        int num = 1;
        int numsize = n*n;
        while (true) {
            for (int i = left; i <= right; ++i) {
                arr[top][i] = num++;
            }
            top++;
            if (num > numsize) break;
            for (int i = top; i <= buttom; ++i) {
                arr[i][right] = num++;

            }
            right--;
            if (num > numsize) break;
            for (int i = right; i >= left; --i) {
                arr[buttom][i] = num++;
            }
            buttom--;
            if (num > numsize) break;
            for (int i = buttom; i >= top; --i) {
                arr[i][left] = num++;
            }
            left++;
            if (num > numsize) break;

        }
        return arr;
    }
}

Python3 Code:

from typing import List
import numpy as np
class Solution:
    def generateMatrix(self, n: int)->List[List[int]]:
        arr = np.array([[0] * n] * n)
        left = 0
        right = n - 1
        top = 0
        buttom = n - 1
        num = 1
        numsize = n * n
        while True:
            for i in range(left, right + 1):
                arr[top][i] = num
                num += 1
            top += 1
            if num > numsize:
                break
            for i in range(top, buttom + 1):
                arr[i][right] = num
                num += 1
            right -= 1
            if num > numsize:
                break
            for i in range(right, left - 1, -1):
                arr[buttom][i] = num
                num += 1
            buttom -= 1
            if num > numsize:
                break
            for i in range(buttom, top - 1, -1):
                arr[i][left] = num
                num += 1
            left += 1
            if num > numsize:
                break
        return arr.tolist()

C++ Code:

class Solution {
public:
    vector<vector<int>> generateMatrix(int n) {
        vector <vector <int>> arr(n, vector <int>(n));
        int left = 0, right = n-1, top = 0, buttom = n - 1, num = 1, numsize = n * n;
        while (true) {
            for (int i = left; i <= right; ++i) {
                arr[top][i] = num++;
            }
            top++;
            if (num > numsize) break;
            for (int i = top; i <= buttom; ++i) {
                arr[i][right] = num++;
            }
            right--;
            if (num > numsize) break;
            for (int i = right; i >= left; --i) {
                arr[buttom][i] = num++;
            }
            buttom--;
            if (num > numsize) break;
            for (int i = buttom; i >= top; --i) {
                arr[i][left] = num++;
            }
            left++;
            if (num > numsize) break;

        }
        return arr;
    }
};

Swift Code:

class Solution {
    func generateMatrix(_ n: Int) -> [[Int]] {
        var arr:[[Int]] = Array.init(repeating: Array.init(repeating: 0, count: n), count: n)
        var left = 0, right = n - 1
        var top = 0, bottom = n - 1
        var num = 1, numSize = n * n

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

        return arr
    }
}

Go Code:

func generateMatrix(n int) [][]int {
    res := make([][]int, n)
    for i := 0; i < n; i++ {
        res[i] = make([]int, n)
    }
    left, right := 0, n - 1
    top, buttom := 0, n - 1
    size, num := n * n, 1
    for {
        for i := left; i <= right; i++ {
            res[top][i] = num
            num++
        }
        top++
        if num > size { break }

        for i := top; i <= buttom; i++ {
            res[i][right] = num
            num++
        }
        right--
        if num > size { break }

        for i := right; i >= left; i-- {
            res[buttom][i] = num
            num++
        }
        buttom--
        if num > size { break }

        for i := buttom; i >= top; i-- {
            res[i][left] = num
            num++
        }
        left++
        if num > size { break }
    }
    return res
}