Question
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"]
Answer
Brute-force的做法是两层for循环，遍历每个string，看有无string可以与之匹配为palindrome。
较好的做法是利用HashMap。
对于String a来说，考虑两种情况：
1. a为空，则a与集合中任意已经对称的string可以组成结果
2. a不为空。
则有三种情况：
a. reverse(a)在集合中
b. a[0..i]对称，且reverse(a[i + 1,...,n])在集合中。如"aaabc", "cba"
c. a[i,..,n]对称，且reverse(a[0,...,i])在集合中。如"abcc", "ba"
public class Solution {
public List<List<Integer>> palindromePairs(String[] words) {
List<List<Integer>> result = new ArrayList<>();
if (words == null || words.length == 0) {
return result;
}
// Record each string position
Map<String, Integer> map = new HashMap<>();
Set<Integer> palindromeSet = new HashSet<>();
int len = words.length;
for (int i = 0; i < len; i++) {
map.put(words[i], i);
if (isPalindrome(words[i])) {
palindromeSet.add(i);
}
}
// Traverse
for (int i = 0; i < len; i++) {
String word = words[i];
if (word.length() == 0) {
for (int index : palindromeSet) {
addResult(result, i, index);
}
}
int l = word.length();
for (int j = 0; j < l; j++) {
String front = word.substring(0, j);
String back = word.substring(j, l);
String rFront = reverse(front);
String rBack = reverse(back);
if (isPalindrome(front) && map.containsKey(rBack)) {
addResult(result, map.get(rBack), i);
}
if (isPalindrome(back) && map.containsKey(rFront)) {
addResult(result, i, map.get(rFront));
}
}
}
return result;
}
private String reverse(String s) {
StringBuilder sb = new StringBuilder(s);
return sb.reverse().toString();
}
private void addResult(List<List<Integer>> result, int start, int end) {
if (start == end) {
return;
}
List<Integer> indexes = new ArrayList<>();
indexes.add(start);
indexes.add(end);
result.add(indexes);
}
private boolean isPalindrome(String s) {
int i = 0;
int j = s.length() - 1;
while (i < j) {
if (s.charAt(i) != s.charAt(j)) {
return false;
}
i++;
j--;
}
return true;
}
}