標籤:word search search leetcode
Word Search
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.
class Solution {public: bool search(vector< vector<char> >& board,int x,int y,string word,vector< vector<bool> >& hasUsed) { if(word.size() == 0)return true; int rows = board.size(),cols = board[0].size(); if(x > 0 && !hasUsed[x-1][y] && board[x-1][y] == word[0])//上方 { hasUsed[x-1][y] = true; if(search(board,x-1,y,word.substr(1),hasUsed))return true; hasUsed[x-1][y] = false; } if(y < cols - 1 && !hasUsed[x][y+1] && board[x][y+1] == word[0])//右方 { hasUsed[x][y+1] = true; if(search(board,x,y+1,word.substr(1),hasUsed))return true; hasUsed[x][y+1] = false; } if(x < rows - 1 && !hasUsed[x+1][y] && board[x+1][y] == word[0])//下方 { hasUsed[x+1][y] = true; if(search(board,x+1,y,word.substr(1),hasUsed))return true; hasUsed[x+1][y] = false; } if(y > 0 && !hasUsed[x][y-1] && board[x][y-1] == word[0])//左方 { hasUsed[x][y-1] = true; if(search(board,x,y-1,word.substr(1),hasUsed))return true; hasUsed[x][y-1] = false; } return false; } bool exist(vector<vector<char> > &board, string word) { int rows = board.size(); if(rows <= 0)return false; int cols = board[0].size(); if(cols <= 0)return false; int i,j; vector< vector<bool> >hasUsed(rows);//用於標記該位置是否走過 for (i = 0;i < rows;i++) { vector<bool> tmp(cols,false); hasUsed[i] = tmp; } for (i = 0;i < rows;i++) { for (j = 0;j < cols;j++) { if(!hasUsed[i][j] && board[i][j] == word[0]) { hasUsed[i][j] = true; if(search(board,i,j,word.substr(1),hasUsed))return true; hasUsed[i][j] = false; } } } return false; }};
leetcode 之 Word Search