标签:turn any img mamicode uil info substr continue key
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
标签:turn any img mamicode uil info substr continue key
原文地址:https://www.cnblogs.com/wentiliangkaihua/p/12258318.html