[LeetCode] 59. Spiral Matrix II 螺旋矩阵 II
Given an integer n, generate a square matrix filled with elements from 1 to n^2 in spiral order.
For example,
Given n = 3,
You should return the following matrix:
[
[ 1, 2, 3 ],
[ 8, 9, 4 ],
[ 7, 6, 5 ]
]
与 54. Spiral Matrix 类似,这次给定一个整数n,以螺旋顺序填充元素到n^2的矩阵。
Java:
class Solution {
public int[][] generateMatrix(int n) {
int[][] res = new int[n][n];
int k = 1;
int top = 0, bottom = n - 1, left = 0, right = n - 1;
while (left < right && top < bottom) {
for (int j = left; j < right; j++) {
res[top][j] = k++;
}
for (int i = top; i < bottom; i++) {
res[i][right] = k++;
}
for (int j = right; j > left; j--) {
res[bottom][j] = k++;
}
for (int i = bottom; i > top; i--) {
res[i][left] = k++;
}
left++;
right--;
top++;
bottom--;
}
if (n % 2 != 0)
res[n / 2][n / 2] = k;
return res;
}
}
Python:
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++:
class Solution {
public:
vector<vector<int> > generateMatrix(int n) {
vector<vector<int> > res(n, vector<int>(n, 1));
int val = 1, p = n;
for (int i = 0; i < n / 2; ++i, p -= 2) {
for (int col = i; col < i + p; ++col)
res[i][col] = val++;
for (int row = i + 1; row < i + p; ++row)
res[row][i + p - 1] = val++;
for (int col = i + p - 2; col >= i; --col)
res[i + p - 1][col] = val++;
for (int row = i + p - 2; row > i; --row)
res[row][i] = val++;
}
if (n % 2 != 0) res[n / 2][n / 2] = val;
return res;
}
};
类似题目:
[LeetCode] 54. Spiral Matrix 螺旋矩阵
All LeetCode Questions List 题目汇总
[LeetCode] 59. Spiral Matrix II 螺旋矩阵 II的更多相关文章
- [LeetCode] 885. Spiral Matrix III 螺旋矩阵之三
On a 2 dimensional grid with R rows and C columns, we start at (r0, c0) facing east. Here, the north ...
- 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 or ...
- LeetCode 59. Spiral Matrix II (螺旋矩阵之二)
Given an integer n, generate a square matrix filled with elements from 1 to n2 in spiral order. For ...
- LeetCode: 59. Spiral Matrix II(Medium)
1. 原题链接 https://leetcode.com/problems/spiral-matrix-ii/description/ 2. 题目要求 给定一个正整数n,求出从1到n平方的螺旋矩阵.例 ...
- Leetcode#59 Spiral Matrix II
原题地址 相比于Spiral Matrix(参见这篇文章)要简单一些,因为是方阵,所以代码简洁一些. 注意当n是奇数的时候,中心小块要单独赋值(代码21行) 代码: vector<vector& ...
- PAT 1105 Spiral Matrix[模拟][螺旋矩阵][难]
1105 Spiral Matrix(25 分) This time your job is to fill a sequence of N positive integers into a spir ...
- [leetcode]59. Spiral Matrix II螺旋遍历矩阵2
Given a positive integer n, generate a square matrix filled with elements from 1 to n^2 in spiral or ...
- LeetCode OJ:Spiral MatrixII(螺旋矩阵II)
Given an integer n, generate a square matrix filled with elements from 1 to n2 in spiral order. For ...
- LeetCode OJ: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 or ...
随机推荐
- BLE——协议层次结构
未完待续…… BLE协议 Bluetooth Application Applications GATT-Based Profiles/Services Bluetooth Core (Stack) ...
- Kotlin反射实践操作详解
继续对反射进行实战. 获取构造方法: 先定义一个主构造方法,2个次构造方法,接下来咱们用反射来获取一下构造方法: 其结果: [fun <init>(kotlin.Int, kotlin.S ...
- Qualification Rounds(Codeforces Round #438 by Sberbank and Barcelona Bootcamp (Div. 1 + Div. 2 combined)+状态压缩)
题目链接 传送门 题意 现总共有\(n\)个题目\(k\)支参赛队伍,已知每个题目各队伍是否会写,现问你能否从题目中选出一个子序列使得每支队伍最多只会写一半的题目. 思路 对于每个题目我们用二进制压缩 ...
- VUE之路
最近研究了下Vue这个前端框架,不得不说这个前端框架很是厉害.不过对于习惯了jQuery的我来说,刚上手那会儿还是踩了很多的坑啊.那会儿觉得天啊,这个Vue框架特别的绕,并且也更复杂.不过待我写了几天 ...
- 清除DNS缓存和刷新DHCP列表
ipconfig /release 只是释放IP地址,然后还需要ipconfig /renew在重新获取一下 如何清除DNS缓存?开始-运行,如下图所示: 在谈出的对话框中输入“cmd”,如下图所示: ...
- docker的daemon配置
文件:/etc/docker/daemon.json,如果没有就创建 修改后重启生效:systemctl restart docker 示例内容: { "registry-mirrors&q ...
- 转发: JS中的call()和apply()方法和区别 --小白变色记
一.方法定义: apply:调用一个对象的一个方法,用另一个对象替换当前对象.例如:B.apply(A, arguments);即A对象应用B对象的方法. call:调用一个对象的一个方法,用另一个对 ...
- Apache Solr < 8.2.0远程命令执行漏洞(CVE-2019-0193)
介绍:Apache Solr 是一个开源的搜索服务器.Solr 使用 Java 语言开发,主要基于 HTTP 和 Apache Lucene 实现. 漏洞原因:此次漏洞出现在Apache Solr的D ...
- Git的个人总结
Git Git简史: 同生活中的许多伟大事物一样,Git 诞生于一个极富纷争大举创新的年代. Linux 内核开源项目有着为数众多的参与者. 绝大多数的 Linux 内核维护工作都花在了提交补丁和保存 ...
- BZOJ 3553: [Shoi2014]三叉神经树 LCT
犯傻了,想到了如果是 0->1 的话就找最深的非 1 编号,是 1 -> 0 的话就找最深的非 0 编号. 但是没有想到这个东西可以直接维护. 假设不考虑叶子节点,那么如果当前点的值是 1 ...