问题
直觉:因为我们必须通过上/下/左/右的方式找到单词数组中存在的单词(在网格/板上)。 可以用回溯来完成遍历 我们可以使用搜索单词 trie,因为这也可以通过检查树上是否有前缀来帮助我们早期识别。这是为了避免不必要的遍历板(即,如果前缀不存在于特里树中,则使用前缀的字符串或单词形式不会出现在特里树中)
方法:我们将所有单词[]放入trie树中,然后遍历棋盘中的每个单元格(i,j),在所有四个方向上形成各种字符串,然后在所有列表中 trie 中间存在的字符串。
class Solution { public List<string> findWords(char[][] board, String[] words) { int max = 0; Trie t = new Trie(); for (String w : words) { t.insert(w); } int arr[][] = new int[board.length][board[0].length]; StringBuilder str = new StringBuilder(); Set<string> list = new HashSet();// to have only unique strings/words for (int i = 0; i (list); } public void traverse(int i, int j, char b[][], int arr[][], int n, int m, Trie t, StringBuilder str,Set<string> list) { str.append(b[i][j]);// add current cell character to form a potential prefix/word if (!t.startWith(str.toString())) {//early checking of prefix before moving forward to avoid un-necessary traversal str.deleteCharAt(str.length() - 1); return; } if (t.present(str.toString())) list.add(str.toString()); arr[i][j] = 1;// mark current cell visited to avoid visiting the same cell the current recursive call stack int dirs[][] = { { 0, -1 }, { 0, 1 }, { -1, 0 }, { 1, 0 } };// left,right,up,down for (int dir[] : dirs) { int I = i + dir[0]; int J = j + dir[1]; if (I = 0 && J >= 0 && arr[I][J] == 0) { traverse(I, J, b, arr, n, m, t, str, list); } } arr[i][j] =0;// mark unvisited str.deleteCharAt(str.length()-1);// remove the last added character } } class Node { Node node[] = new Node[26]; boolean flag; public boolean isPresent(char c) { return node[c - 'a'] != null; } public Node get(char c) { return node[c - 'a']; } public void add(char c, Node n) { node[c - 'a'] = n; } public void setFlag() { this.flag = true; } public boolean getFlag() { return this.flag; } } class Trie { Node root; public Trie() { root = new Node(); } public void insert(String s) { Node node = root; for (int i = 0; i </string></string></string>
以上是单词搜索 有关II的详细信息,请关注图灵教育的其他相关文章!