leetcode 212. Word Search II

标签：为我   dex   array   map   continue   cto   使用   child   etc   

一眼看出是使用回溯法。

function findWords(board, words) {
     var rows = board.length,  cols = board[0].length, res = []
      var visited = new Array(rows).fill(0).map(function (el, i) {
        return new Array(cols).fill(false)
      })
      var set = new Set();
      for (var word of words) {
        set.add(word);
      }

      function backtrack(cur, x, y) {
        if (x == rows  || x < 0 || y == cols  || y < 0 || visited[x][y]) {
          return;
        }

        var c = board[x][y];
        if (set.has(cur + c)) {
          res.push(cur + c);
          set.delete(cur + c);
        }
        visited[x][y] = true;
        backtrack(cur + c, x + 1, y);
        backtrack(cur + c, x - 1, y);
        backtrack(cur + c, x, y + 1);
        backtrack(cur + c, x, y - 1);
        visited[x][y] = false;
      }

      for (var i = 0; i < rows; i++) {
        for (var j = 0; j < cols;  j++) {
          backtrack("", i, j);
        }
      }
      return res;
    }

于是超时了。因为我们没有对字符的方向进行控制，有的字符只能向上或向下，这是由words数组里面的单词决定的。因此我们可以弄一个前缀树，这样我们就能找到字符间的流向了。

 function Node(value) {
      this.value = value
      this.word = null
      this.children = new Array(26)
    }
    class Tire {
      constructor() {
        this.root = new Node(null)
      }
      addWord(word) {
        var node = this.root;
        for (var c of word) {
          var index = c.charCodeAt(0) - 97
          if (!node.children[index]) {
            node.children[index] = new Node(c)
          }
          node = node.children[index]
        }
        node.word = word
      }
    }



    function findWords(board, words) {
      var rows = board.length, cols = board[0].length;
      var visited = new Array(rows).fill(0).map(function (el, i) {
        return new Array(cols).fill(false)
      })
      var trie = new Tire()
      for (let word of words) {
        trie.addWord(word)
      }
      var root = trie.root, res = []
      //用于简洁地取得下一个trie节点
      function getChild(root, i, j) {
        return root.children[board[i][j].charCodeAt(0) - 97]
      }
      function backtrack(node, x, y) {
        if (x == rows || x < 0 || y == cols || y < 0 || visited[x][y]) {
          return;
        }
        var child = getChild(node, x, y)
        if (!child) {
          return
        }
        if (child.word !== null) {
          res.push(child.word);
          child.word = null
        }
        visited[x][y] = true;
        backtrack(child, x + 1, y);
        backtrack(child, x - 1, y);
        backtrack(child, x, y + 1);
        backtrack(child, x, y - 1);
        visited[x][y] = false;
      }

      for (var i = 0; i < rows; i++) {
        for (var j = 0; j < cols; j++) {
          if (getChild(root, i, j) == null) {
            continue
          }
          backtrack(root, i, j)
        }
      }
      return res;
    }

leetcode 212. Word Search II

标签：为我   dex   array   map   continue   cto   使用   child   etc   

原文地址：https://www.cnblogs.com/rubylouvre/p/12105069.html