Ubuntu手记
windows简单是简单,好用是好用,就是不知道哪里被我捣鼓坏了,vscode的内置终端一直用不了,这对于我这个vscode爱好者来说能忍?
根本忍不了啊
而在使用了Ubuntu写代码后发现这玩意儿比我以前好使太多了,反正也就写代码的时候会切到这个系统上,所以再难用也没啥问题,因此准备入坑。
想着像写github手记来当做指令小抄一样方便平日里的使用,特意写了一份Ubuntu命令手记,时不时更新。
常用命令显示当前文件夹路径
pwd
跳向文件夹的命令
cd
显示当前目录所有文件或文件夹
ls
ls -h:所有子文件与子文件夹(包括隐藏
ls -l:列表方式展示文件详细信息
ls -a:以人性化方式展示文件大侠
* 即可以代替任何符号数字
\?代表一个数值或符号
若 ls 1[1-6]5.py 代表显示当前文件夹中带有.py后缀,且第一个为1,第二个为1到6的数字,第三个为5的 所有文件。
总之正则啦
将ls命令输出的结果保存到一个指定文件中去。一个大于号会覆盖之前文件中的东西,两个大于号在文件中末尾加入。
ls
......
然后
ls > or >> fi ...
LeetCode.1143.583.72.322.343(动态规划)
1143.最长公共子序列class Solution { public: int longestCommonSubsequence(string text1, string text2) { int text1S = text1.size(); int text2S = text2.size(); vector<vector<int>> dp(text1S + 1, vector(text2S + 1, 0)); for (int i = text1S - 1; i >= 0; i--) { for (int j = text2S - 1; j >= 0; j--) { if (text1[i] == text2[j]) { dp[i][j] = dp[i + 1][j + 1] + 1; } dp[i][j] = max(dp[i + 1][j], dp[i][j]); dp[i] ...
LeetCode.300.673(动态规划)
代码随想录就是我爹,没他这两题这个月都不能AC
今天两题全靠评论区AC,下次一定要雪耻
300.最长递增子序列class Solution { public: int lengthOfLIS(vector<int> &nums) { if (nums.size() <= 1) return nums.size(); vector<int> dp(nums.size(), 1); int result = 0; for (int i = 1; i < nums.size(); i++) { for (int j = 0; j < i; j++) { if (nums[i] > nums[j]) dp[i] = max(dp[i], dp[j] + 1); } resu ...
LeetCode.5.413.91.139(动态规划)
5.最长回文子串class Solution { public: /* int sumRemainder(string a) { int sum = 0; for (int i = 0; i < a.length(); i++) sum = (sum + a[i]) % 123; return sum; } bool judge(string a) { for (int len = a.length() / 2; len >= 1; len /= 2) for (int i = 0; i < a.length() / 2 / len; i++) if (sumRemainder(a.substr(0 + i * len, len)) != sumRemainder(a.substr(a.length() - (i + 1) * len, len))) ...
LeetCode.213.55.45.62(动态规划)
213.打家劫舍-iiclass Solution { public: int rob(vector<int> &nums) { int n = nums.size(); /* if (n <= 1) { return nums[0]; } int Max = 0; for (int i = 0; i < n; i++) { vector<int> path(n, 0); path[0] = nums[i]; path[1] = max(nums[(i + 1) % n], path[0]); int pathMax = path[1]; for (int j = i + 2; j < i - 1 + n; j++) { int q = 0; ...
LeetCode.17.22.79(递归&回溯)
17.电话号码的字母组合class Solution { public: vector<string> ans; vector<string> dictionary = {"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}; string path; vector<string> letterCombinations(string digits) { if (digits.empty()) return {}; find(digits, 0); return ans; } void find(string digits, int pos) { ...
LeetCode.78.90.47.39.40(递归&回溯)
回溯法模板78.子集class Solution { public: vector<vector<int>> subsets(vector<int> &nums) { vector<vector<int>> ans; ans.push_back({}); for (int i = 0; i < nums.size(); i++) add(nums, ans, i, {nums[i]}, nums.size()); return ans; } void add(vector<int> &nums, vector<vector<int>> &ans, int pos, vector<int> temp, int n) { ans.push_back(temp); ...
LeetCode.1091.130.797(广度优先搜索&深度优先搜索)
今天的题目真滴难,一天都花在这上面,JAVA都没碰
1091.二进制矩阵中的最短路径class Solution { public: /* int shortestPathBinaryMatrix(vector<vector<int>> &grid) { for (int i = 0; i < grid.size(); i++) { for (int j = 0; j < grid[0].size(); j++) { if (grid[i][j] == 1) grid[i][j] = 0; else grid[i][j] = INT_MAX; } } if (grid[0][0] == 0 || grid[grid.size() - 1][grid[0].s ...
LeetCode.200.547.117.572(广度优先搜索&深度优先搜索)
200.岛屿数量class Solution { public: int numIslands(vector<vector<char>> &grid) { int sum = 0; for (int i = 0; i < grid.size(); i++) { for (int j = 0; j < grid[0].size(); j++) { if (grid[i][j] == '1') { count(grid, j, i); sum++; } } } return sum; } void count(vector<vector<char>> &g ...
LeetCode.438.715.209(滑动窗口)
438.找到字符串中所有字母异位词class Solution { public: vector<int> findAnagrams(string s, string p) { if (p.length() > s.length()) return {}; vector<int> result; vector<int> hashS(26, 0); vector<int> hashP(26, 0); for (int i = 0; i < p.length(); i++) { hashP[p[i] - 'a']++; } for (int i = 0; i < s.length(); i++) { hashS[s[i] - 'a']++; ...