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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,"SEE", -> returns true,"ABCB", -> returns false.
Array Backtracking
#include <iostream> #include <vector> #include <string> using namespace std; class Solution { public: bool exist(vector<vector<char> > &board, string word) { if(word.length()<1) return true; if(board.size()==0||board[0].size()==0) return false; for(int i =0;i<board.size();i++){ for(int j =0;j<board[0].size();j++){ if(board[i][j]==word[0]&&helpFun(board,word,1,i,j)) return true; } } return false; } bool helpFun(vector<vector<char> >&board,string & word,int idx,int beg_i,int beg_j) { if(idx == word.size()) return true; char tmp = board[beg_i][beg_j]; board[beg_i][beg_j] = ‘*‘; if(beg_i>0&&board[beg_i-1][beg_j]==word[idx]&&helpFun(board,word,idx+1,beg_i-1,beg_j)){ board[beg_i][beg_j] = tmp; return true; } if(beg_i<board.size()-1&&board[beg_i+1][beg_j]==word[idx]&&helpFun(board,word,idx+1,beg_i+1,beg_j)){ board[beg_i][beg_j] = tmp; return true; } if(beg_j>0&&board[beg_i][beg_j-1]==word[idx]&&helpFun(board,word,idx+1,beg_i,beg_j-1)){ board[beg_i][beg_j] = tmp; return true; } if(beg_j<board[0].size()-1&&board[beg_i][beg_j+1]==word[idx]&&helpFun(board,word,idx+1,beg_i,beg_j+1)){ board[beg_i][beg_j] = tmp; return true; } board[beg_i][beg_j] = tmp; return false; } }; int main() { vector<vector< char> > board{{‘A‘,‘B‘,‘C‘,‘E‘},{‘S‘,‘F‘,‘C‘,‘S‘},{‘A‘,‘D‘,‘E‘,‘E‘}}; Solution sol; cout<<sol.exist(board,"ABCB")<<endl; return 0; }
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原文地址：http://www.cnblogs.com/Azhu/p/4356543.html
