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:

Input: ["abcd","dcba","lls","s","sssll"]
Output: [[0,1],[1,0],[3,2],[2,4]] 
Explanation: The palindromes are ["dcbaabcd","abcddcba","slls","llssssll"]

Example 2:

Input: ["bat","tab","cat"]
Output: [[0,1],[1,0]] 
Explanation: The palindromes are ["battab","tabbat"]

public List<List<Integer>> palindromePairs(String[] words) {
    List<List<Integer>> res = new ArrayList<List<Integer>>();
    if(words == null || words.length == 0){
        return res;
    }
    //build the map save the key-val pairs: String - idx
    HashMap<String, Integer> map = new HashMap<>();
    for(int i = 0; i < words.length; i++){
        map.put(words[i], i);
    }
    
    //special cases: "" can be combine with any palindrome string
    if(map.containsKey("")){
        int blankIdx = map.get("");
        for(int i = 0; i < words.length; i++){
            if(isPalindrome(words[i])){
                if(i == blankIdx) continue;
                res.add(Arrays.asList(blankIdx, i));
                res.add(Arrays.asList(i, blankIdx));
            }
        }
    }
    
    //find all string and reverse string pairs
    for(int i = 0; i < words.length; i++){
        String cur_r = reverseStr(words[i]);
        if(map.containsKey(cur_r)){
            int found = map.get(cur_r);
            if(found == i) continue;
            res.add(Arrays.asList(i, found));
        }
    }
    
    //find the pair s1, s2 that 
    //case1 : s1[0:cut] is palindrome and s1[cut+1:] = reverse(s2) => (s2, s1)
    //case2 : s1[cut+1:] is palindrome and s1[0:cut] = reverse(s2) => (s1, s2)
    for(int i = 0; i < words.length; i++){
        String cur = words[i];
        for(int cut = 1; cut < cur.length(); cut++){
            if(isPalindrome(cur.substring(0, cut))){
                String cut_r = reverseStr(cur.substring(cut));
                if(map.containsKey(cut_r)){
                    int found = map.get(cut_r);
                    if(found == i) continue;
                    res.add(Arrays.asList(found, i));
                }
            }
            if(isPalindrome(cur.substring(cut))){
                String cut_r = reverseStr(cur.substring(0, cut));
                if(map.containsKey(cut_r)){
                    int found = map.get(cut_r);
                    if(found == i) continue;
                    res.add(Arrays.asList(i, found));
                }
            }
        }
    }
    
    return res;
}

public String reverseStr(String str){
    StringBuilder sb= new StringBuilder(str);
    return sb.reverse().toString();
}

public 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;
}

4种情况，1：空字符“” + 任何palindrome

2： s1和s2互为reverse，这种情况也是一次循环就可以，因为有s1+s2和s2+s1两种情况

3. 4. 如上面所述