Leetcode---Word Search

最新推荐文章于 2020-10-27 20:35:59 发布

原创最新推荐文章于 2020-10-27 20:35:59 发布 · 393 阅读

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本文介绍了如何在二维网格中搜索指定单词，采用深度优先搜索（DFS）算法，确保单词由相邻单元格构成且不重复使用同一字母单元格。

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.

Have you been asked this question in an interview?

无他，DFS耳。

int depth;
bool dfs(vector<vector<char>> &board, string word,int i,int j,vector<vector<bool>> &m){
depth++;
int len=word.length();
if(i<0 || i>=board.size() || j<0 || j>=board[0].size()){
depth--;
return false;
}
if((depth==len) && (board[i][j]==word[depth-1])){
depth--;
return true;
}
if(depth>word.length()){
depth--;
return false;
}
if(board[i][j]!=word[depth-1]){
depth--;
return false;
}
if(i-1>=0 && i-1<board.size() && j>=0 && j<board[0].size() ){
if(m[i-1][j]==false ){
m[i-1][j]=true;
if(dfs(board,word,i-1,j,m)==true){
depth--;
return true;
}
m[i-1][j]=false;
}
}
if(i+1>=0 && i+1<board.size() && j>=0 && j<board[0].size()){
if(m[i+1][j]==false){
m[i+1][j]=true;
if(dfs(board,word,i+1,j,m)==true){
depth--;
return true;
}
m[i+1][j]=false;
}
}
if(i>=0 && i<board.size() && j-1>=0 && j-1<board[0].size()){
if(m[i][j-1]==false){
m[i][j-1]=true;
if(dfs(board,word,i,j-1,m)==true){
depth--;
return true;
}
m[i][j-1]=false;
}
}
if(i>=0 && i<board.size() && j+1>=0 && j+1<board[0].size()){
if(m[i][j+1]==false){
m[i][j+1]=true;
if(dfs(board,word,i,j+1,m)==true){
depth--;
return true;
}
m[i][j+1]=false;
}
}
depth--;
return false;
}
bool exist(vector<vector<char> > &board, string word) {
vector<vector<bool>> m(board.size(),vector<bool>(board[0].size(),false));
for(int i=0;i<board.size();i++){
for(int j=0;j<board[0].size();j++){
if(board[i][j]==word[0]){
m[i][j]=true;
if(dfs(board,word,i,j,m))
return true;
m[i][j]=false;
}
}
}
return false;
}