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

.

题目解析：

深度优先算法，找到一个合适的字符，从左上右下依次的遍历，直到找到所有的word字符为止。

bool dfs(vector<vector<char> > &board, int x, int y, string word)
{
if(word.size() == 0)return true;

bool flag = false;
if(x-1>=0 && board[x-1][y] == word[0])
{
board[x-1][y] = '#';
flag = dfs(board, x-1, y, word.substr(1));
board[x-1][y] = word[0];
}

if(!flag && y-1>=0 && board[x][y-1] == word[0])
{
board[x][y-1] = '#';
flag = dfs(board, x, y-1, word.substr(1));
board[x][y-1] = word[0];
}

if(!flag && x+1<board.size() && board[x+1][y] == word[0])
{
board[x+1][y] = '#';
flag = dfs(board, x+1, y, word.substr(1));
board[x+1][y] = word[0];
}

if(!flag && y+1<board[0].size() && board[x][y+1] == word[0])
{
board[x][y+1] = '#';
flag = dfs(board, x, y+1, word.substr(1));
board[x][y+1] = word[0];
}

return flag;
}
bool exist(vector<vector<char> > &board, string word) {
// Note: The Solution object is instantiated only once.
if(word.size() < 1)return true;
int row = board.size();
int col = board[0].size();
set<string> st;
for(int i = 0; i < row; i++)
for(int j = 0; j < col; j++)
if(board[i][j] == word[0])
{
board[i][j] = '#';
if(dfs(board,i,j,word.substr(1)))
return true;
board[i][j] = word[0];
}
return false;
}

方案二：

这个解题中用到dx,dy很好求解二维数组遍历过程。将整个四个分过程整合到一起

const int dx[] = {0,0,1,-1};
const int dy[] = {1,-1,0,0};
class Solution {
public:
unordered_set<long long> flag;
bool check(vector<vector<char> > & board , string& word , int x , int y , int pos){
if(pos == word.size()) return true;
for(int i = 0 ; i < 4 ; i++){
int tx = x + dx[i];
int ty = y + dy[i];
if(tx >= 0 && tx < board.size() && ty >= 0 && ty < board[tx].size()){
if(flag.find(tx*100000+ty) == flag.end() && board[tx][ty] == word[pos]){
flag.insert(tx*100000+ty);
if(check(board , word , tx , ty , pos + 1)){
return true;
}else{
flag.erase(tx*100000+ty);
}
}
}
}
return false;
}
bool exist(vector<vector<char> > &board, string word) {
if(word == "") return true;
int size = board.size();

for(int i = 0 ; i < size ; i++){
for(int j = 0 ; j < board[i].size() ; j++){
if(board[i][j] == word[0]){
flag.clear();
flag.insert(i*100000 + j);
if(check(board , word , i , j , 1)) return true;
}
}
}
return false;
}
};