leetcode-Word Search

leetcode-Word Search

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.

Discuss

典型的DFS題目。Coding過程中注意狀態的變化以及邊界情況。

class Solution {
public:
    bool exist(vector > &board, string word) {
        if (word == "") {
            return false;
        }
        int nNumberOfRows = board.size();
        vector > bVisit;
        for (int i = 0; i < nNumberOfRows; ++i) {
            int nNumberOfCols = board[i].size();
            bVisit.push_back(vector());
            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 > &board, int row, int col, vector > &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 > &board, vector > &bVisit, int row, int col) {
        if (row >= 0 && row < board.size() && 
            col >= 0 && col < board[row].size() &&
            !bVisit[row][col]) {
            return true;
        }
        return false;
    }
};