leetcode 56: Word Search-优快云博客

本文介绍了一个算法，用于在一个二维字符网格中查找指定的单词。该算法通过深度优先搜索（DFS）遍历网格，检查单词是否能按顺序由相邻的字符组成。

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Word SearchApr 18 '12

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.

uncompleted

class Position{
public:
    int x;
    int y;
    Position(int xx, int yy) : x(xx),y(yy){};
    
};

class Solution {
public:
    bool exist(vector<vector<char> > &board, string word) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(word.size() ==0) return true;
        if(board.size()==0) return false;
        Solution::m = board.size();
        Solution::n = board[0].size();
        vector<bool> temp(n,false);
        vector<vector<bool> > flags(m, temp);
        
        for( int i=0; i<m; i++) {
            for( int j=0; j<n; j++) {
                if(board[i][j]==word[0]){
                    flags[i][j] = true;
                    if( dfs( board, word, flags, 0, Position(i, j) ) ) return true;
                    flags[i][j] = false;
                }
            }
        }
        return false;
    }
    
private:
    int m;
    int n;
    
    bool dfs( vector<vector<char> > &board, string & word, vector<vector<bool> >& flags, int level,Position p ) {
        if( level == word.size()-1) return true;
        if( word[level] != board[p.x][p.y]) return false;
        for( int i=-1; i<=1; i++) {
            for( int j=-1; j<=1; j++) {
                if(i+j!=1 && i+j!=-1) continue;
                if( p.x+i>=0 && p.y+j>=0 && p.x+i<m && p.y+j<n && !flags[p.x+i][p.y+j] ) {
                    flags[p.x+i][p.y+j] = true;
                    if( dfs(board, word, flags, level+1, Position( p.x+i, p.y+j) ) )
                        return true;
                    else
                        flags[p.x+i][p.y+j] == false;
                }
            }
        }
        return false;
    }
};

转载于:https://www.cnblogs.com/xishibean/archive/2013/01/27/2951332.html