LeetCode 79. Word Search [python3/C++]

最新推荐文章于 2022-10-14 22:00:00 发布

原创最新推荐文章于 2022-10-14 22:00:00 发布 · 320 阅读

0 ·

CC 4.0 BY-SA版权

本文介绍了一个经典的深度优先搜索算法问题，即在一个二维网格中查找指定的单词。文章详细解析了算法思路并提供了Python及C++两种语言的实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

转载请注明出处: https://blog.youkuaiyun.com/yyying2016/article/details/80174304

一题目

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.

二分析

题目的大意是在一个二维矩阵中，找图中是否有一条路径，使得路径上的字符按顺序连起来是一个目标字符串word。一个点只能经过一次，并且下一个点只能是其上下左右相邻的点。

这是一道典型的深搜题目，建立一个相同维度大小的isvisit二维数组，用来标记每一点是否已经过。对所有点作为起点遍历一遍，如果越界或者字符不满足字符串则return false，如果能到达终点则return true。每次一条路径尝试失败后要重新将isvisit标记为未经过。

三代码

python3：

class Solution:
    
    def exist(self, board, word):
        """
        :type board: List[List[str]]
        :type word: str
        :rtype: bool
        """
        
        dx, dy = [0, 0, 1, -1], [-1, 1, 0, 0];

        def dfs(board, word, k, x, y, isvisit):
            if x < 0 or x >= len(board) or y < 0 or y >= len(board) or isvisit[x][y] == 1 or board[x][y] != word[k]:
                return False;
            if len(word) == k + 1:
                return True;
            isvisit[x][y] = 1
            for i in range(4):
                if dfs(board, word, k + 1, x + dx[i], y + dy[i], isvisit) == True:
                    return True
            isvisit[x][y] = 0
            return False
    
        n = len(board)
        m = len(board[0])
        isvisit =  [[0 for col in range(m)] for row in range(n)]
        for i in range(n):
            for j in range(m):
                isvisit[i][j] = 0
        for i in range(n):
            for j in range(m):
                if dfs(board, word, 0, i, j, isvisit) == True:
                    return True
        return False

C++：

class Solution {
public:
    int dx[4] = {0, 0, 1, -1}, dy[4] = {-1, 1, 0, 0};
    
    bool dfs(vector<vector<char> >& board, string word, int k, int x, int y, vector<vector<int> >& isvisit) {
        if(x < 0 || x >= board.size() || y < 0 || y >= board[0].size() || isvisit[x][y] == 1 || board[x][y] != word[k])
            return false;
        if(word.size() == k + 1)
            return true;
        isvisit[x][y] = 1;
        for(int i = 0; i < 4; i++) {
            if(dfs(board, word, k + 1, x + dx[i], y + dy[i], isvisit) == true)
                return true;
        }
        isvisit[x][y] = 0;
        return false;
    }
    
    bool exist(vector<vector<char>>& board, string word) {
        int n = board.size(), m = board[0].size();
        vector<vector<int> > isvisit(n, vector<int>(m,0));
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                isvisit[i][j] = 0;
            }
        }
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                if(dfs(board, word, 0, i, j, isvisit))
                    return true;
            }
        }
        return false;
    }
};