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【LeetCode】Palindrome Pairs(336)

时间:2016-04-03 15:59:40      阅读:189      评论:0      收藏:0      [点我收藏+]

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1. Description

  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"]

2. Answer  

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;
        }
        //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;
    } 
}

 

【LeetCode】Palindrome Pairs(336)

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原文地址:http://www.cnblogs.com/leesf456/p/5349413.html

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