标签：ati rds bin unique att careful reverse hash hat
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"]
Naive Solution: Time: O(n^2*k) with n the total number of words in the "words" array and k the average length of each word: check each combination see if it‘s palindrome. TLE of course.
Better Solution: Time: O(n*k^2)
think of a word A which contains two part,
1. A = XX + B, XX is palindrome, then "B_reverse + XX + B" will make a palindrome, find if B_reverse exists in the the list
2. A = C + XX , then "C + XX + C_reverse" will make a palindrome, find if C_reverse exists in the list,
To ensure quick search, use HashMap
Be careful about duplicate search: [abcd, dcba],
in first iteration, we look at word abcd, at iteration where sndHalf == "", we add {0,1}
in second iteration, we look at word dcba, at iteration where fstHaf == "", we also add {0, 1}, duplicates
public class Solution {
public List<List<Integer>> palindromePairs(String[] words) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
if (words==null || words.length==0) return res;
Map<String, Integer> map = new HashMap<>();
for (int i=0; i<words.length; i++) {
map.put(words[i], i);
}
for (int i=0; i<words.length; i++) {
int len = words[i].length();
for (int j=0; j<=words[i].length(); j++) {
String fstHalf = words[i].substring(0, j);
String sndHalf = words[i].substring(j);
if (isPalindrome(fstHalf)) {
String sndHalfRev = new StringBuffer(sndHalf).reverse().toString();
if (map.containsKey(sndHalfRev) && map.get(sndHalfRev)!=i) { //"aaaa" case
ArrayList<Integer> item = new ArrayList<Integer>();
item.add(map.get(sndHalfRev));
item.add(i);
res.add(new ArrayList<Integer>(item));
}
}
if (isPalindrome(sndHalf)) {
String fstHalfRev = new StringBuffer(fstHalf).reverse().toString();
if (map.containsKey(fstHalfRev) && map.get(fstHalfRev)!=i && sndHalf.length()!=0) {
ArrayList<Integer> item = new ArrayList<Integer>();
item.add(i);
item.add(map.get(fstHalfRev));
res.add(new ArrayList<Integer>(item));
}
}
}
}
return res;
}
public boolean isPalindrome(String str) {
int r = str.length()-1;
int l = 0;
while (l <= r) {
if(str.charAt(l++) != str.charAt(r--)) return false;
}
return true;
}
}
标签：ati rds bin unique att careful reverse hash hat
原文地址：http://www.cnblogs.com/apanda009/p/7825450.html
