Leetcode: Anagrams

来源：互联网发布：招商银行贵金属软件编辑：程序博客网时间：2024/06/07 02:22

Given an array of strings, return all groups of strings that are anagrams.

Note: All inputs will be in lower-case.

题意理解有些困惑，其实是返回变位词个数大于等于2的所有单词，全部放到一个vector里面去。

然后就比较简单了，编程珠玑里面有提到。

class Solution {public:    vector<string> anagrams(vector<string> &strs) {        vector<string> result;        if (strs.size() <= 1) {            return result;        }                vector<pair<string, string>> strnodes;        string sorted;        for (int i = 0; i < strs.size(); ++i) {            sorted = strs[i];            sort(sorted.begin(), sorted.end());            strnodes.push_back(make_pair(strs[i], sorted));        }        sort(strnodes.begin(), strnodes.end(), anagram_less);                int count = 1;        sorted = strnodes[0].second;        int i = 1;        for (; i < strnodes.size(); ++i) {            if (strnodes[i].second == sorted) {                ++count;            }            else {                if (count >= 2) {                    for (int j = i - count; j < i; ++j) {                        result.push_back(strnodes[j].first);                    }                }                sorted = strnodes[i].second;                count = 1;            }        }        if (count >= 2) {            for (int j = i - count; j < i; ++j) {                result.push_back(strnodes[j].first);            }        }                return result;    }        static bool anagram_less(pair<string, string> p1, pair<string, string> p2) {        return p1.second < p2.second;    }};

================第二次=================

class Solution {public:    vector<string> anagrams(vector<string> &strs) {        unordered_map<string, multiset<string>> groups;        for (int i = 0; i < strs.size(); ++i) {            string sorted = strs[i];            sort(sorted.begin(), sorted.end());            auto iter = groups.find(sorted);            if (iter == groups.end()) {                multiset<string> group;                group.insert(strs[i]);                groups[sorted] = group;            }            else {                iter->second.insert(strs[i]);            }        }                vector<string> result;        for (auto iter = groups.begin(); iter != groups.end(); ++iter) {            if (iter->second.size() >= 2) {                result.insert(result.end(), iter->second.begin(), iter->second.end());            }        }                return result;    }};

貌似Leetcode里面不要求回文字符串相邻，可以进一步简化一些：

class Solution {public:    vector<string> anagrams(vector<string> &strs) {        vector<string> result;        unordered_map<string, int> groups;        for (int i = 0; i < strs.size(); ++i) {            string sorted = strs[i];            sort(sorted.begin(), sorted.end());            auto iter = groups.find(sorted);            if (iter == groups.end()) {                groups.insert({sorted, i});            }            else {                result.push_back(strs[i]);                if (iter->second >= 0) {                    result.push_back(strs[iter->second]);                    iter->second = -1;                }            }        }                return result;    }};

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