给定一组唯一的单词, 找出所有不同 的索引对(i, j),使得列表中的两个单词, words[i] + words[j]
,可拼接成回文串。
示例 1:
输入: ["abcd","dcba","lls","s","sssll"]
输出: [[0,1],[1,0],[3,2],[2,4]]
解释: 可拼接成的回文串为 ["dcbaabcd","abcddcba","slls","llssssll"]
示例 2:
输入: ["bat","tab","cat"]
输出: [[0,1],[1,0]]
解释: 可拼接成的回文串为 ["battab","tabbat"]
来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/palindrome-pairs 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。
class Solution {
public:
vector<vector<int>> palindromePairs(vector<string>& words) {
unordered_map<string, int> w_id;
set<int> wdLen;
for(int i = 0; i < words.size(); ++i)
{
w_id[words[i]] = i;//字符串idx
wdLen.insert(words[i].size());//字符串长度
}
vector<vector<int>> ans;
string front, back, revword;
for(int i = 0; i < words.size(); ++i)
{
revword = words[i];//逆序的字符串
reverse(revword.begin(),revword.end());
if(w_id.count(revword) && w_id[revword] != i)
ans.push_back({i, w_id[revword]});//字符串的逆序存在
//遍历words[i]可能的子串长度,寻找前部分存在或者后部分存在
//且自身剩余的子串为回文
int len = words[i].size();
for(auto it = wdLen.begin(); *it < len; ++it)
{
front = words[i].substr(0, *it);
reverse(front.begin(),front.end());
back = words[i].substr(*it);
if(w_id.count(front) && ispalind(back))//前缀的逆存在
ans.push_back({i, w_id[front]});
}
for(auto it = wdLen.begin(); *it < len; ++it)
{
front = revword.substr(0, *it);
back = revword.substr(*it);
if(w_id.count(front) && ispalind(back))//后缀的逆存在
ans.push_back({w_id[front], i});
}
}
return ans;
}
bool ispalind(string& s)
{
int l = 0, r = s.size()-1;
while(l < r)
if(s[l++] != s[r--])
return false;
return true;
}
};
904 ms 45.6 MB
class trie
{
public:
unordered_map<char, trie*> next;
int suffix = -1;
void insert(string& s, int idx)
{
trie *cur = this;
for(int i = s.size()-1; i >= 0; --i)//单词逆序插入
{
if(!cur->next[s[i]])
cur->next[s[i]] = new trie();
cur = cur->next[s[i]];
}
cur->suffix = idx;//结束时记录单词编号
}
};
class Solution {
public:
vector<vector<int>> palindromePairs(vector<string>& words) {
trie * t = new trie(), *cur;
vector<vector<int>> ans;
string revword;
for(int i = 0; i < words.size(); ++i)
{
t->insert(words[i], i);
}
for(int i = 0; i < words.size(); ++i)
{
int n = words[i].size(), j, k;
cur = t;
for(j = 0; j < n; ++j)
{
if(cur->suffix != -1 && cur->suffix != i
&& ispalind(words[i], j, n-1))//单词的前缀的逆序在trie中,剩余的为回文
ans.push_back({i, cur->suffix});
if(!cur->next[words[i][j]])
break;
cur = cur->next[words[i][j]];
}
for(j = 0; j <= n; ++j)//等号上下只取一次,否则答案有重复的
{ // j == n 时包含了完整字符串的情况
cur = t;
for(k = n-j; k < n; ++k)//遍历单词的后缀
{
if(!cur->next[words[i][k]])
break;
cur = cur->next[words[i][k]];
}
if(k==n && cur->suffix != -1
&& cur->suffix != i
&& ispalind(words[i], 0, n-j-1))//该后缀的逆在trie中,且前部分为回文
ans.push_back({cur->suffix, i});
}
}
return ans;
}
bool ispalind(string s, int l, int r)
{
while(l < r)
if(s[l++] != s[r--])
return false;
return true;
}
};
940 ms 141.3 MB
trie 改用数组 trie* next[26] = {NULL};
提高运行效率
280 ms 208.5 MB