标签：list i++ ice add question rom user case 问题
问题描述：
Given a list of unique words, find all pairs of distinct indices (i, j) in the given list, so that the concatenation of the two words, i.e. words[i] + words[j] is a palindrome.
Example 1:
Given words = ["bat", "tab", "cat"]
Return [[0, 1], [1, 0]]
The palindromes are ["battab", "tabbat"]
Example 2:
Given words = ["abcd", "dcba", "lls", "s", "sssll"]
Return [[0, 1], [1, 0], [3, 2], [2, 4]]
The palindromes are ["dcbaabcd", "abcddcba", "slls", "llssssll"]
Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.
解题思路：
我一开始的想法是把所有单词根据其最后一个字符存入hash表，构成一个<char, vector<string>>的map
然后再遍历每一个单词，根据他们第一个单词在hash表中寻找，构成新的字符串，来判断是否为回文，若是则要加入。
但是这个方法耗时很大，若n为数组长度，k为字符串平聚长度。 应为n*(n-1)*2k
还有一种方法是O(n*k2) n 为数组长度，而k为字符串的平均长度。
这里需要额外注意的是！
字符串数组中！有没有肯恩含有空数组！！！
代码：
我的：
class Solution { public: vector<vector<int>> palindromePairs(vector<string>& words) { vector<vector<int>> ret; unordered_map<string, int> m; unordered_map<char, vector<string>> endWith; bool emptyExist = false; for(int i = 0; i < words.size(); i++){ m[words[i]] = i; if(words[i] != "") endWith[words[i].back()].push_back(words[i]); else emptyExist = true; } for(int i = 0; i < words.size(); i++){ if(emptyExist && words[i] != ""){ if(isPalindrome(words[i])){ ret.push_back({m[""], i}); ret.push_back({i, m[""]}); } } for(auto str : endWith[words[i][0]]){ string newS = words[i] + str; if(m[str] != i && isPalindrome(newS)){ ret.push_back({i, m[str]}); } } } return ret; } bool isPalindrome(string s){ int l = 0, r = s.size()-1; while(l < r){ if(s[l++] != s[r--]) return false; } return true; } };
快的：
class Solution { public: vector<vector<int>> palindromePairs(vector<string>& words) { vector<vector<int>> ret; unordered_map<string, int> dict; for(int i = 0; i < words.size(); i++){ string w = words[i]; reverse(w.begin(), w.end()); dict[w] = i; } if(dict.count("") != 0){ for(int i = 0; i < words.size(); i++){ if(words[i] != ""){ if(isPalindrome(words[i])) ret.push_back({dict[""], i}); } } } for(int i = 0; i < words.size(); i++){ for(int j = 0; j < words[i].size(); j++){ string left = words[i].substr(0, j); string right = words[i].substr(j, words[i].size() - j); if(dict.count(left) != 0 && isPalindrome(right) && dict[left] != i) ret.push_back({i, dict[left]}); if(dict.count(right) != 0 && isPalindrome(left) && dict[right] != i) ret.push_back({dict[right], i}); } } return ret; } bool isPalindrome(string s){ int l = 0, r = s.size()-1; while(l < r){ if(s[l++] != s[r--]) return false; } return true; } };
标签：list i++ ice add question rom user case 问题
原文地址：https://www.cnblogs.com/yaoyudadudu/p/9345387.html
