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.
Example:
board = [ ['A','B','C','E'], ['S','F','C','S'], ['A','D','E','E'] ] Given word = "ABCCED", return true. Given word = "SEE", return true. Given word = "ABCB", return false.
使用dfs
1 class Solution {//dfs my 2 public boolean exist(char[][] board, String word) { 3 if(null==board||0==board.length){ 4 return false; 5 } 6 int row = board.length; 7 int col = board[0].length; 8 boolean[][] flag = new boolean[row][col]; 9 return dfs(board,flag,0,0,row,col,word,0); 10 } 11 private boolean dfs(char[][] board, boolean[][] flag,int x,int y,int row,int col,String word,int index){ 12 if(index>=word.length()){ 13 return true; 14 } 15 if(x<0||y<0||x>=row||y>=col||flag[x][y]){ 16 return false; 17 } 18 char c = word.charAt(index); 19 boolean re = false; 20 if(0==index){ 21 for (int i = 0; i < row; i++) { 22 for (int j = 0; j < col; j++) { 23 if(board[i][j]==c){ 24 flag[i][j]=true; 25 re = dfs(board,flag,i+1,j,row,col,word,index+1)||dfs(board,flag,i-1,j,row,col,word,index+1)||dfs(board,flag,i,j+1,row,col,word,index+1)||dfs(board,flag,i,j-1,row,col,word,index+1); 26 flag[i][j]=false; 27 } 28 if(re){ 29 return true; 30 } 31 } 32 } 33 } 34 else{ 35 if(board[x][y]==c){ 36 flag[x][y]=true; 37 re= dfs(board,flag,x+1,y,row,col,word,index+1)||dfs(board,flag,x-1,y,row,col,word,index+1)||dfs(board,flag,x,y+1,row,col,word,index+1)||dfs(board,flag,x,y-1,row,col,word,index+1); 38 flag[x][y]=false; 39 } 40 } 41 return re; 42 } 43 }
简洁写法
1 class Solution {//dfs my 2 public boolean exist(char[][] board, String word) { 3 if(null==board||0==board.length){ 4 return false; 5 } 6 int row = board.length; 7 int col = board[0].length; 8 boolean[][] flag = new boolean[row][col]; 9 char c = word.charAt(0); 10 for (int i = 0; i < row; i++) { 11 for (int j = 0; j < col; j++) { 12 if(board[i][j]==c){ 13 flag[i][j]=true; 14 if(dfs(board,flag,i+1,j,word,1)||dfs(board,flag,i-1,j,word,1)||dfs(board,flag,i,j+1,word,1)||dfs(board,flag,i,j-1,word,1)){ 15 return true; 16 } 17 flag[i][j]=false; 18 } 19 } 20 } 21 return false; 22 } 23 private boolean dfs(char[][] board, boolean[][] flag,int x,int y,String word,int index){ 24 if(index>=word.length()){ 25 return true; 26 } 27 if(x<0||y<0||x>=board.length||y>=board[0].length||flag[x][y]){ 28 return false; 29 } 30 boolean re = false; 31 if(board[x][y]==word.charAt(index)){ 32 flag[x][y]=true; 33 re= dfs(board,flag,x+1,y,word,index+1)||dfs(board,flag,x-1,y,word,index+1)||dfs(board,flag,x,y+1,word,index+1)||dfs(board,flag,x,y-1,word,index+1); 34 flag[x][y]=false; 35 } 36 return re; 37 } 38 }
进阶题
单词搜索 LeetCode212 https://www.cnblogs.com/zhacai/p/10641592.html