回文构词法,将字母顺序打乱.可将字母重新排序,若它们相等,则属于同一组anagrams. 可通过hashmap来做,将排序后的字母作为key.注意后面取hashmap值时的做法. vector<string> anagrams(vector<string> &strs) { unordered_map<string, vector<string>> group; for (const auto &s : strs) { string key…

Given an array of strings, group anagrams together. Example: Input: ["eat", "tea", "tan", "ate", "nat", "bat"], Output: [ ["ate","eat","tea"], ["nat","…

LeetCode第28题 Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack. Example 1: Input: haystack = "hello", needle = "ll" Output: 2 Example 2: Input: haystack = "aaaaa", needle…

If you understand the comments below, never will you make mistakes with binary search! thanks to A simple CPP solution with lower_bound and C++ O(logn) Binary Search that handles duplicate, thanks to phu1ku 's answer on the second post. http://en.cpp…

Anagrams Given an array of strings, return all groups of strings that are anagrams. Note: All inputs will be in lower-case.     Anagrams:即字母个数和字母都相同,但是字母顺序不相同的词   e.g. "tea","and","ate","eat","dan".   retu…

Given an array of strings, return all groups of strings that are anagrams. Note: All inputs will be in lower-case. anagrams 的意思是两个词用相同的字母组成  比如 “dog" "god" 思路: 把单词排序 如 dog 按字母排序变为 dgo 用unordered_map<string, int> 记录排序后序列第一次出现时,字符串在输入st…

Given an array of strings, return all groups of strings that are anagrams. Note: All inputs will be in lower-case. 解题思路:首先要理解,什么是anagrams,ie.“tea”.“tae”.“aet”,然后就十分好做了,new一个hashmap,使用一个排过序的String作为key,重复了就往里面添加元素,这里出现一个小插曲,就是排序的时候不要用PriorityQueue,因为P…

原题地址 Anagram:变位词.两个单词是变位词关系的条件是:组成单词的字符相同,只是顺序不同 第一次看这道题看了半天没明白要干嘛,丫就不能给个样例输入输出么..后来还是看网上其他人的总结知道是怎么回事. 通常的做法是:把字符串内的字符排序,这样,凡是变位词都会变成相同的单词.用map记录这样的单词出现了几个,如果超过1个,则加入结果集中. 代码: vector<string> anagrams(vector<string> &strs) { vector<stri…

Given an array of strings, return all groups of strings that are anagrams. Note: All inputs will be in lower-case. 他的意思就是回文构词法,即单词里的字母的种类和数目没有改变,仅仅是改变了字母的排列顺序. input= ["abc", "bca", "bac", "bbb", "bbca", &…

Given an array of strings, return all groups of strings that are anagrams. Note: All inputs will be in lower-case. 题解: 判断字符串是否为回文构词法生成的.找出所有由同一回文构词法生成的字符串对. 使用map用于散列. 将strs中的字符串strs[i],在串内进行字典排序,生成key,原始s[i]不变. 将该字符串s[i]映射到key所对应位置.map[key].push_bac…