***(leetcode_backtracking) Word Search

Word Search

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Question Solution 

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.

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解题思路： 类似于迷宫，递归回溯。需要一个辅助数组记录走过的位置，防止同一个位置被使用多次

class Solution {public:    bool exist(vector<vector<char> > &board, string word) {        if(word.length()==0)            return true;                vector<vector<bool> > visited;        vector<bool> tmp;        int i, j;        for(i=0; i < board.size(); i++){            for(j=0; j < board[0].size(); j++){                tmp.push_back(false);            }            visited.push_back(tmp);        }                for(i=0; i < board.size(); i++){            for(j=0; j < board[0].size(); j++){                if(board[i][j]==word[0]){                    visited[i][j] = true;                    if(word.length()==1 || search(board, i, j, word.substr(1,word.length()-1), visited) ){                        return true;                    }                    visited[i][j] = false;                }            }        }        return false;    }        bool search(vector<vector<char> > &board, int i, int j, string word, vector<vector<bool> > visited){        if(word.length()==0)            return true;        int direction[][2] = {{0,1}, {0,-1}, {1,0}, {-1,0}};                for (int k=0; k < 4; k++) {          int ii = i + direction[k][0];          int jj = j + direction[k][1];          if ( ii>=0 && ii<board.size() && jj>=0 && jj<board[i].size() && board[ii][jj]==word[0] && visited[ii][jj]==false) {              visited[ii][jj] = true;              if (word.length()==1 || search(board,ii,jj,word.substr(1,word.length()-1),visited)) {                  return true;              }              visited[ii][jj] = false;          }      }      return false;      }};