336. Palindrome Pairs

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"]

摘自
https://discuss.leetcode.com/topic/40657/150-ms-45-lines-java-solution/2
跟wordladders一样的思想，字串很多，逐个来就是N^2，利用已知的生成再配合hashmap检索能加快速度。

public List<List<Integer>> palindromePairs(String[] words) {
List<List<Integer>> ret = new ArrayList<>();
if (words == null || words.length < 2) return ret;
Map<String, Integer> map = new HashMap<String, Integer>();
for (int i=0; i<words.length; i++) map.put(words[i], i);
for (int i=0; i<words.length; i++) {
// System.out.println(words[i]);
for (int j=0; j<=words[i].length(); j++) { // notice it should be "j <= words[i].length()"
String str1 = words[i].substring(0, j);
String str2 = words[i].substring(j);
if (isPalindrome(str1)) {
String str2rvs = new StringBuilder(str2).reverse().toString();
if (map.containsKey(str2rvs) && map.get(str2rvs) != i) {
List<Integer> list = new ArrayList<Integer>();
list.add(map.get(str2rvs));
list.add(i);
ret.add(list);
// System.out.printf("isPal(str1): %s\n", list.toString());
}
}
if (isPalindrome(str2)) {
String str1rvs = new StringBuilder(str1).reverse().toString();
// check "str.length() != 0" to avoid duplicates
if (map.containsKey(str1rvs) && map.get(str1rvs) != i && str2.length()!=0) {
List<Integer> list = new ArrayList<Integer>();
list.add(i);
list.add(map.get(str1rvs));
ret.add(list);
// System.out.printf("isPal(str2): %s\n", list.toString());
}
}
}
}
return ret;
}

private boolean isPalindrome(String str) {
int left = 0;
int right = str.length() - 1;
while (left <= right) {
if (str.charAt(left++) !=  str.charAt(right--)) return false;
}
return true;
}

The 

<=

 in 

for
(int j=0; j<=words[i].length(); j++)

 is aimed to handle empty string in the input. Consider the test case of ["a", ""];

Since we now use 

<=

 in 

for
(int j=0; j<=words[i].length(); j++)

 instead of 

<

. There may be duplicates
in the output (consider test case ["abcd", "dcba"]). Therefore I put a 

str2.length()!=0

 to
avoid duplicates.