来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/scramble-string 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。 2. 解题 2.1 记忆化递归 按照长度 1 到 n-1 把 s1 , s2 切开,递归判断 注意用哈希表记录下来一些已经得到的结果,否则超时195 / 285 个通过测试用例 代...
https://leetcode.com/problems/minimum-depth-of-binary-tree/ https://leetcode.com/problems/populating-next-right-pointers-in-each-node/ https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/ https://leetcode.com/problems/binary-tree-right-side-view/ https://leetcode.c...
1616.Split-Two-Strings-to-Make-Palindrome (M+) 1754.Largest-Merge-Of-Two-Strings (M+) 1849.Splitting-a-String-Into-Descending-Consecutive-Values (M+) 2468.Split-Message-Based-on-Limit (H-) Abbreviation 408.Valid-Word-Abbreviation (M) 411.Minimum-Unique-Word-Abbreviation (H) 527.Word-Abbre...
//Time Limit ExceededclassSolution {public:boolisScramble(strings1,strings2) {if(s1 == s2)returntrue;stringstr1 = s1, str2 =s2; sort(str1.begin(), str1.end()); sort(str2.begin(), str2.end());if(str1 != str2)returnfalse;for(inti =1; i < s1.size(); ++i) {strings11 =...
class Solution { public String addStrings(String num1, String num2) { int i = num1.length() - 1, j = num2.length() - 1, add = 0; StringBuffer ans = new StringBuffer(); while (i >= 0 || j >= 0 || add != 0) { int x = i >= 0 ? num1.charAt(i) - '0' : ...
211 211. Add and Search Word - Data structure design.java Medium [Backtracking, Design, Trie] O(n) to search and to add word < O(mn), depends on the input. m = # of words Java 439 43 43. Multiply Strings.java Medium [Math, String] O(mn) O(mn) Java 440 621 621. Task Sched...
https://leetcode.com/problems/greatest-common-divisor-of-strings/discuss/303759/My-4-Lines-C%2B%2B-Solution https://leetcode.com/problems/greatest-common-divisor-of-strings/discuss/303781/JavaPython-3-3-codes-each%3A-Recursive-iterative-and-regex-w-brief-comments-and-analysis. ...
// Solution 1: Use disjoint set, set the node p as the root of an orphan tree. All process nodes on that tree shall beterminated. 代码1 //Code 1 583 Delete Operation for Two Strings // #583 两字符串的删除操作 描述:给定两个字符串,允许你每次从某串删除一个字符。问至少经过多少操作,可...
2. Sliding Window (Arrays, Strings, Hash Tables) Sliding Window covers most of the problems related to top data structures like Arrays, Strings, and HashTables. 3. Tree and Graph Depth First Search (Matrix Traversal) Most Trees and Graphs problems can be solved using Depth First Search (DFS...
Problems List in there Depth First Search Problems List in there Breadth First Search Problems List in there Binary Search <img alt=""> 二分搜索的经典写法。需要注意的三点: 循环退出条件,注意是 low <= high,而不是 low < high。 mid 的取值,mid := low + (high-low)>>1 low 和 high 的...