7、螺旋矩阵2
大约 5 分钟数据结构算法算法基地面试刷题
题目描述
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
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]
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;
}
}
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
}
