LeetCode (22) Word Search

题目描述

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.

For example,

Given board =

word = “ABCCED”, -> returns true,

word = “SEE”, -> returns true,

word = “ABCB”, -> returns false.

本题要求给定一个数组，判断数组中是否存在某字符串。本题可以使用回溯法去解决。给定一张表，用于标记每个元素是否遍历过，若查找的元素与字符串中相应元素相同，则查找下一个元素，若不相同，则返回false，并从上一个元素开始重新查找。

直接上代码：

class Solution {
public:
    bool findWord(vector<vector<char> > &board, string word, int i, int j,
                  int k, vector<vector<bool> > &finded)
    {
        if (k == word.size())
            return true;
        if (i < 0 || j < 0 || i >= board.size() || j >= board[0].size()
            || finded[i][j] || board[i][j] != word[k])
            return false;
        finded[i][j] = true;
        bool flag = findWord(board, word, i - 1, j, k + 1, finded)
                 || findWord(board, word, i + 1, j, k + 1, finded)
                 || findWord(board, word, i, j - 1, k + 1, finded)
                 || findWord(board, word, i, j + 1, k + 1, finded);
        finded[i][j] = false;
        return flag;
    }

    bool exist(vector<vector<char> > &board, string word) {
        if (!board.size() && word.empty())
            return true;
        else if (!board.size())
            return false;

        int row = board.size();
        int col = board[0].size();
        vector<vector<bool> > finded;
        for (int i = 0; i != row; ++i)
        {
            vector<bool> f;
            for (int j = 0; j != col; ++j)
            {
                f.push_back(false);
            }
            finded.push_back(f);
        }

        for (int i = 0; i != row; ++i)
        {
            for (int j = 0; j != col; ++j)
            {
                if (findWord(board, word, i, j, 0, finded))
                    return true;
            }
        }

        return false;
    }
};