12. 矩阵中的路径

来源

https://leetcode-cn.com/problems/ju-zhen-zhong-de-lu-jing-lcof/

题目

请设计一个函数，用来判断在一个矩阵中是否存在一条包含某字符串所有字符的路径。路径可以从矩阵中的任意一格开始，每一步可以在矩阵中向左、右、上、下移动一格。如果一条路径经过了矩阵的某一格，那么该路径不能再次进入该格子。例如，在下面的 3×4 的矩阵中包含一条字符串“bfce”的路径（路径中的字母用加粗标出）。

[["a","b","c","e"],
["s","f","c","s"],
["a","d","e","e"]]

但矩阵中不包含字符串“abfb”的路径，因为字符串的第一个字符 b 占据了矩阵中的第一行第二个格子之后，路径不能再次进入这个格子。

示例:

示例 1.

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
输出：true

示例 2.

输入：board = [["a","b"],["c","d"]], word = "abcd"
输出：false

提示：

• 1 <= board.length <= 200
• 1 <= board[i].length <= 200

思路

图，深度优先，回溯法

解题

/**
 * @param {character[][]} board
 * @param {string} word
 * @return {boolean}
 */
var exist = function (board, word) {
  const rows = board.length;
  const cols = board[0].length;
  const visited = new Array(rows).fill(0).map(() => new Array(cols).fill(0));
  // word 指针
  let pathIndex = 0;
  // 递归
  function existCore(i, j) {
    // word 已找到
    if (pathIndex === word.length) {
      return true;
    }
    let has = false;
    // 条件
    if (
      i >= 0 &&
      i < rows &&
      j >= 0 &&
      j < cols &&
      board[i][j] === word[pathIndex] &&
      !visited[i][j]
    ) {
      pathIndex++;
      // 标记
      visited[i][j] = 1;
      // 上下左右查找下一层
      has =
        existCore(i - 1, j) ||
        existCore(i, j - 1) ||
        existCore(i, j + 1) ||
        existCore(i + 1, j);
      // 回溯
      if (!has) {
        pathIndex--;
        visited[i][j] = 0;
      }
    }
    return has;
  }
  // 遍历
  for (var i = 0; i < rows; i++) {
    for (var j = 0; j < cols; j++) {
      if (existCore(i, j)) return true;
    }
  }
  return false;
};

解题 2

用原数组进行标记，空间复杂度小

/**
 * @param {character[][]} board
 * @param {string} word
 * @return {boolean}
 */
var exist = function (board, word) {
  const rows = board.length;
  const cols = board[0].length;
  let pathIndex = 0;
  function existCore(i, j) {
    if (pathIndex === word.length) {
      return true;
    }
    let has = false;
    if (
      i >= 0 &&
      i < rows &&
      j >= 0 &&
      j < cols &&
      board[i][j] === word[pathIndex]
    ) {
      pathIndex++;
      let tmp = board[i][j];
      board[i][j] = "-";
      has =
        existCore(i - 1, j) ||
        existCore(i, j - 1) ||
        existCore(i, j + 1) ||
        existCore(i + 1, j);
      if (!has) {
        pathIndex--;
        board[i][j] = tmp;
      }
    }
    return has;
  }
  for (var i = 0; i < rows; i++) {
    for (var j = 0; j < cols; j++) {
      if (existCore(i, j)) return true;
    }
  }
  return false;
};

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