classSolution{public:vector<int>printNumbers(intn){if(n==0)return{};vector<int> ans; n =pow(10, n);stringcur ="0";for(inti=1; i<n; i++){ cur = add(cur,"1"); ans.push_back(stoi(cur)); }returnans; }stringadd(stringcur,stringone){intcarry =0;inti = cur.size()-1;int...
(nSum + '0');//需要强转 takeOver = 1; } }else{ number[i] = (char)('0' + nSum);//需要强转 break; } } return isOverflow; } private static void printNumber(char[] number){ boolean isBeginning0 = true; int len = number.length; for(int i = 0; i < len; i++){ if(...
#链接:https://leetcode-cn.com/problems/palindrome-number/solution/jing-xin-hui-zong-python3de-5chong-shi-xian-fang-f/classSolution:# 方法一:将int转化成str类型:双向队列 # 复杂度:O(n^2)[每次pop(0)都是O(n)..比较费时]defisPalindrome(x:int)->bool:lst=list(str(x))whilelen(lst)>1:...
😎1 合并两个有序数组 给你两个按 非递减顺序 排列的整数数组 nums1 和 nums2,另有两个整数 m 和 n ,分别表示 nums1 和 nums2 中的元素数目。 请你 合并 nums2 到 nums1 中,使合并后的数组同样按 非递减顺序 排列。 注意:最终,合并后数组不应由函数返回,而是存储在数组 nums1 中。为了应对这种情...
solve() def solve(self): """Solve the n queens puzzle and print the number of solutions""" positions = [-1] * self.size self.put_queen(positions, 0) print("Found", self.solutions, "solutions.") def put_queen(self, positions, target_row): """ Try to place a queen on target_...
encode-n-ary-tree-to-binary-tree encrypt-and-decrypt-strings equal-row-and-column-pairs equal-sum-arrays-with-minimum-number-of-operations er-cha-shu-de-jing-xiang-lcof er-cha-shu-de-shen-du-lcof er-cha-shu-de-zui-jin-gong-gong-zu-xian-lcof er-cha-shu-ran-se-UGC er-...
str = 'wenjtop' print (str) # 输出字符串 print (str[0:-1]) # 输出第一个到倒数第二个的所有字符 print (str[0]) # 输出字符串第一个字符 print (str[2:5]) # 输出从第三个开始到第五个的字符 print (str[2:]) # 输出从第三个开始的后的所有字符 print (str * 2) # 输出字符串两...
2870-minimum-number-of-operations-to-make-array-empty 2966-divide-array-into-arrays-with-max-difference 2971-find-polygon-with-the-largest-perimeter Longest Palindrome in a String - GFG Next Greater Element - GFG Pattern Printing - GFG Print 1 To N Without Loop - GFG Print GFG n...
class Solution { public boolean isMatch(String s, String p) { int m = s.length() + 1, n = p.length() + 1; boolean[][] dp = new boolean[m][n]; dp[0][0] = true; for(int j = 2; j < n; j += 2){ dp[0][j] = dp[0][j - 2] && p.charAt(j-1) == '*'...
0590 N-ary Tree Postorder Traversal N叉树的后序遍历 LeetCode 力扣 Python CSDN Easy 二叉树 0605 Can Place Flowers LeetCode 力扣 Python CSDN Easy 贪心 0606 Construct String from Binary Tree 根据二叉树创建字符串 LeetCode 力扣 Python CSDN Easy 二叉树 0623 Add One Row to Tree 在二叉树中增加一行...