LeetCode 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"]

题意：给出一个字符串数组，将任意两个拼凑起来为对称字符串的索引对输出
思路：刚开始直接按照题意写超时。后来改成哈希查找通过

如4种情况：1、如果字符串有为空的，则任意一个对称字符串与之可以拼凑

2、如果一个字符串是另一个字符串的反转，则可以拼凑

3、如果字符串s1的前n个字符是对称的，即s1[0,n），并且s1[n,len)与s2是互为相反的，则(s2,s1)构成对称字符串

4、如果字符串s2的后几个字符是对称的，即s1[n,len),并且s1[0,n)与s2是互为相反的，则(s1, s2)构成对称字符串

代码如下：
class Solution
{
private String strrev(String s)
{
StringBuilder sb = new StringBuilder(s);
return sb.reverse().toString();
}

private boolean palindrome(String s)
{
int len = s.length();
for (int i = 0; i < len / 2; i++)
{
if (s.charAt(i) != s.charAt(len - 1 - i))
{
return false;
}
}

return true;
}

public List<List<Integer>> palindromePairs(String[] words)
{
List<List<Integer>> ret = new ArrayList<List<Integer>>();

if (null == words || 0 == words.length)
{
return ret;
}

Map<String, Integer> hs = new HashMap<String, Integer>();
for (int i = 0; i < words.length; i++)
{
hs.put(words[i], i);
}

if (hs.containsKey(""))
{
int index = hs.get("");
for (int i = 0; i < words.length; i++)
{
if (palindrome(words[i]))
{
if (i == index) continue;
ret.add(Arrays.asList(i, index));
ret.add(Arrays.asList(index, i));
}
}
}

for (int i = 0; i < words.length; i++)
{
String rev = strrev(words[i]);
if (hs.containsKey(rev))
{
int index = hs.get(rev);
if (index == i) continue;
ret.add(Arrays.asList(i, index));
}
}

for (int i = 0; i < words.length; i++)
{
for (int len = 1; len < words[i].length(); len++)
{
if (palindrome(words[i].substring(0, len)))
{
String rev = strrev(words[i].substring(len));
if (hs.containsKey(rev))
{
int index = hs.get(rev);
if (i == index) continue;
ret.add(Arrays.asList(index, i));
}
}

if (palindrome(words[i].substring(len)))
{
String rev = strrev(words[i].substring(0, len));
if (hs.containsKey(rev))
{
int index = hs.get(rev);
if (i == index) continue;
ret.add(Arrays.asList(i, index));
}
}
}
}
return ret;
}
}