给定一组 互不相同 的单词, 找出所有 不同 的索引对 (i, j),使得列表中的两个单词, words[i] + words[j] ,可拼接成回文串。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/palindrome-pairs
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import java.util.*; class Solution { Map<String, Integer> indices = new HashMap<String, Integer>(); public List<List<Integer>> palindromePairs(String[] words) { int n = words.length; for (int i = 0; i < n; ++i) { indices.put(new StringBuffer(words[i]).reverse().toString(), i); } List<List<Integer>> ret = new ArrayList<List<Integer>>(); for (int i = 0; i < n; i++) { String word = words[i]; int m = words[i].length(); for (int j = 0; j <= m; j++) { if (isPalindrome(word, j, m - 1)) { int leftId = findWord(word, 0, j - 1); if (leftId != -1 && leftId != i) { ret.add(Arrays.asList(i, leftId)); } } if (j != 0 && isPalindrome(word, 0, j - 1)) { int rightId = findWord(word, j, m - 1); if (rightId != -1 && rightId != i) { ret.add(Arrays.asList(rightId, i)); } } } } return ret; } public boolean isPalindrome(String s, int left, int right) { while (left < right) { if (s.charAt(left++) != s.charAt(right--)) { return false; } } return true; } public int findWord(String s, int left, int right) { return indices.getOrDefault(s.substring(left, right + 1), -1); } }
import java.util.ArrayList; import java.util.Arrays; import java.util.List; class Solution { private static String enrich(String word) { StringBuffer sb = new StringBuffer(); sb.append('$'); for (int i = 0; i < word.length(); ++i) { sb.append(word.charAt(i)).append('$'); } return sb.toString(); } private static boolean[][] getPalindromePrefixAndSuffix(String word) { if (word == null || word.length() == 0) { return new boolean[2][0]; } boolean[][] ret = new boolean[2][word.length()]; word = enrich(word); int[] p = new int[word.length()]; int c = -1; int r = -1; // 0 1 2 3 4 5 6 // # 1 # 1 # 1 # for (int i = 0; i < word.length(); ++i) { p[i] = i > r ? 1 : Math.min(p[2 * c - i], r - i + 1); while (i + p[i] < word.length() && i - p[i] >= 0 && word.charAt(i + p[i]) == word.charAt(i - p[i])) { p[i]++; } if (i + p[i] - 1 > r) { r = i + p[i] - 1; c = i; } if (i != 0 && i != word.length() - 1) { if (i - p[i] + 1 == 0) { ret[0][(i + p[i] - 2) >> 1] = true; } if (i + p[i] == word.length()) { ret[1][(i - p[i] + 2) >> 1] = true; } } } return ret; } /** * O(n * m) * * @param words * @return */ private static List<List<Integer>> solve(String[] words) { List<List<Integer>> ret = new ArrayList<>(); Trie prefixTrie = new Trie(); Trie suffixTrie = new Trie(); for (int i = 0; i < words.length; ++i) { prefixTrie.insert(words[i], i); suffixTrie.insert(new StringBuilder(words[i]).reverse().toString(), i); } for (int i = 0; i < words.length; ++i) { String word = words[i]; boolean[][] palindromePrefixAndSuffix = getPalindromePrefixAndSuffix(word); boolean[] prefix = palindromePrefixAndSuffix[0]; boolean[] suffix = palindromePrefixAndSuffix[1]; int[] suffixIndex = suffixTrie.query(word); int[] prefixIndex = prefixTrie.query(new StringBuilder(word).reverse().toString()); int m = word.length(); /** * 是否存在word完全逆序的字符串 */ if (suffixIndex[m] != -1 && suffixIndex[m] != i) { ret.add(Arrays.asList(i, suffixIndex[m])); } for (int j = 0; j < m; ++j) { /** * word字符串 [0, j] 为回文串 * 需要 [j+1, m) 的字符串逆序后拼接到前面 */ if (prefix[j]) { if (prefixIndex[m - j - 1] != -1 && prefixIndex[m - j - 1] != i) { ret.add(Arrays.asList(prefixIndex[m - j - 1], i)); } } /** * word字符串 [j, m)为回文串 * 需要 [0, j-1] 的字符串逆序后拼接到后面 */ if (suffix[j]) { if (suffixIndex[j] != -1 && suffixIndex[j] != i) { ret.add(Arrays.asList(i, suffixIndex[j])); } } } } return ret; } public static List<List<Integer>> palindromePairs(String[] words) { if (words == null || words.length == 0) { return new ArrayList<>(0); } return solve(words); } } class Trie { private Node root; static class Node { Node[] children = new Node[26]; int index = -1; } public Trie() { this.root = new Node(); } public void insert(String word, int index) { Node p = root; for (int i = 0; i < word.length(); ++i) { int path = word.charAt(i) - 'a'; if (p.children[path] == null) { p.children[path] = new Node(); } p = p.children[path]; } p.index = index; } public int[] query(String word) { Node p = root; int[] ret = new int[word.length() + 1]; Arrays.fill(ret, -1); ret[0] = root.index; for (int i = 0; i < word.length(); ++i) { int path = word.charAt(i) - 'a'; if (p.children[path] == null) { break; } p = p.children[path]; ret[i + 1] = p.index; } return ret; } }