zoukankan      html  css  js  c++  java
  • 0079. Word Search (M)

    Word Search (M)

    题目

    Given a 2D board and a word, find if the word exists in the grid.

    The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.

    Example:

    board =
    [
      ['A','B','C','E'],
      ['S','F','C','S'],
      ['A','D','E','E']
    ]
    
    Given word = "ABCCED", return true.
    Given word = "SEE", return true.
    Given word = "ABCB", return false.
    

    题意

    判断给定矩阵中是否存在一组相邻字符序列,能够组成目标字符串。即以矩阵中一字符为起点,只能上下左右走且不能重复访问元素,判断有无这样一条路径,使起点到终点的字符序列正好组成目标字符串。

    思路

    回溯法。


    代码实现

    Java

    class Solution {
        // 控制左上右下四个方向坐标的变化
        private int[] iPlus = {0, -1, 0, 1};
        private int[] jPlus = {-1, 0, 1, 0};
        private int m, n;
    
        public boolean exist(char[][] board, String word) {
            m = board.length;
            n = board[0].length;
            boolean[][] visited = new boolean[m][n];
    
            // 先找到与第一个字符相同的位置,以该位置为起点搜索路径
            for (int i = 0; i < m; i++) {
                for (int j = 0; j < n; j++) {
                    if (board[i][j] == word.charAt(0)) {
                        visited[i][j] = true;
                        if (search(board, i, j, word, 0, visited)) {
                            return true;
                        }
                        visited[i][j] = false;
                    }
                }
            }
            
            return false;
        }
    
        private boolean search(char[][] board, int i, int j, String word, int index, boolean[][] visited) {
            if (index == word.length() - 1) {
                return true;
            }
    
            for (int x = 0; x < 4; x++) {
                int nextI = i + iPlus[x];
                int nextJ = j + jPlus[x];
    
                // 判断邻接点是否在矩阵内
                if (nextI < 0 || nextI >= m || nextJ < 0 || nextJ >= n) {
                    continue;
                }
    
                // 如果邻接点与字符串中下一个字符匹配,且尚未被访问,则继续递归搜索
                if (!visited[nextI][nextJ] && board[nextI][nextJ] == word.charAt(index + 1)) {
                    visited[nextI][nextJ] = true;
                    if (search(board, nextI, nextJ, word, index + 1, visited)) {
                        return true;
                    }
                    visited[nextI][nextJ] = false;
                }
            }
            
            return false;
        }
    }
    

    JavaScript

    /**
     * @param {character[][]} board
     * @param {string} word
     * @return {boolean}
     */
    var exist = function (board, word) {
      let m = board.length
      let n = board[0].length
      let visited = new Array(m).fill(0).map(v => [])
      for (let i = 0; i < m; i++) {
        for (let j = 0; j < n; j++) {
          if (board[i][j] === word[0] && dfs(board, m, n, i, j, word, 1, visited)) {
            return true
          }
        }
      }
      return false
    }
    
    let dfs = function (board, m, n, i, j, word, index, visited) {
      if (index === word.length) {
        return true
      }
      visited[i][j] = true
      let iPlus = [0, -1, 0, 1]
      let jPlus = [-1, 0, 1, 0]
      for (let x = 0; x < 4; x++) {
        let nextI = i + iPlus[x]
        let nextJ = j + jPlus[x]
        if (
          isValid(m, n, nextI, nextJ) &&
          !visited[nextI][nextJ] &&
          board[nextI][nextJ] === word[index] &&
          dfs(board, m, n, nextI, nextJ, word, index + 1, visited)
        ) {
          return true
        }
      }
      visited[i][j] = false
      return false
    }
    
    let isValid = function (m, n, i, j) {
      return i >= 0 && i < m && j >= 0 && j < n
    }
    
  • 相关阅读:
    flex属性导图
    html代码换行造成空格间距问题
    iconfont作用在css伪类中的写法
    JS模态框 简单案例
    JS实时获取输入框中的值
    JS封装addClass、removeClass
    特效 左右滑动轮播图jQuery思路
    JS 字符串两边截取空白的trim()方法的封装
    JavaScript易混淆知识点小回顾--数组方法与字符串方法;
    用GitHub来展示前端页面
  • 原文地址:https://www.cnblogs.com/mapoos/p/13359212.html
Copyright © 2011-2022 走看看