首页 > 代码库 > leetcode-Word Search

leetcode-Word Search

2024-07-13 21:05:03   225人阅读

Word Search

 Total Accepted: 11535 Total Submissions: 58659My Submissions

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 =

[
  ["ABCE"],
  ["SFCS"],
  ["ADEE"]
]
word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.

Have you been asked this question in an interview? 

Discuss


典型的DFS题目。Coding过程中注意状态的变化以及边界情况。

class Solution {
public:
    bool exist(vector<vector<char> > &board, string word) {
        if (word == "") {
            return false;
        }
        int nNumberOfRows = board.size();
        vector<vector<bool> > bVisit;
        for (int i = 0; i < nNumberOfRows; ++i) {
            int nNumberOfCols = board[i].size();
            bVisit.push_back(vector<bool>());
            for (int j = 0; j < nNumberOfCols; ++j) {
                bVisit[i].push_back(false);
            }
        }
        
        for (int i = 0; i < nNumberOfRows; ++i) {
            int nNumberOfCols = board[i].size();
            for (int j = 0; j < nNumberOfCols; ++j) {
                if (board[i][j] == word[0]) {
                    if (dfs(board, i, j, bVisit, word, 1)) {
                        return true;
                    }
                }
            }
        }
        return false;
    }
protected:
    bool dfs(vector<vector<char> > &board, int row, int col, vector<vector<bool> > &bVisit, string &word, int n) {
        if (n == word.size()) {
            return true;
        }
        bVisit[row][col] = true;
        static int dir[][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
        for (int i = 0; i < 4; ++i) {
            int newRow = row + dir[i][0];
            int newCol = col + dir[i][1];
            if (isValid(board, bVisit, newRow, newCol) && board[newRow][newCol] == word[n]) {
                if (dfs(board, newRow, newCol, bVisit, word, n + 1)) {
                    return true;
                }
            }
        }
        bVisit[row][col] = false;
        return false;
    }
    bool isValid(vector<vector<char> > &board, vector<vector<bool> > &bVisit, int row, int col) {
        if (row >= 0 && row < board.size() && 
            col >= 0 && col < board[row].size() &&
            !bVisit[row][col]) {
            return true;
        }
        return false;
    }
};


联系
我们