LeetCode刷题【Array】 Word Search

题目：

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 =

[  ['A','B','C','E'],  ['S','F','C','S'],  ['A','D','E','E']]

word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.

解决方法一：Runtime: 14 ms

public class Solution {    public boolean exist(char[][] board, String word) {if(null==board) return false;if(null==word) return true;int [][] roadMap = new int[board.length][];for(int i=0;i<board.length;i++){roadMap[i] = new int[board[i].length];}for(int i=0;i<board.length;i++){for(int j=0;j<board[i].length;j++){if(board[i][j]==word.charAt(0)){roadMap[i][j] = 1;boolean flag = backTraceWord(board,i,j,roadMap,word,1);if(flag==true) return true;roadMap[i][j] = 0;}}}        return false;    }private boolean backTraceWord(char[][] board, int r,int c, int[][] roadMap, String word, int index){if(index>=word.length()) return true;boolean flag = false;if(c-1>=0&&roadMap[r][c-1]!=1&&board[r][c-1]==word.charAt(index)){roadMap[r][c-1]=1;flag|=backTraceWord(board,r,c-1,roadMap,word,index+1);if(flag==true) return true;roadMap[r][c-1]=0;}if(r-1>=0&&roadMap[r-1][c]!=1&&board[r-1][c]==word.charAt(index)){roadMap[r-1][c]=1;flag|=backTraceWord(board,r-1,c,roadMap,word,index+1);if(flag==true) return true;roadMap[r-1][c]=0;}if(c+1<roadMap[r].length&&roadMap[r][c+1]!=1&&board[r][c+1]==word.charAt(index)){roadMap[r][c+1]=1;flag|=backTraceWord(board,r,c+1,roadMap,word,index+1);if(flag==true) return true;roadMap[r][c+1]=0;}if(r+1<roadMap.length&&roadMap[r+1][c]!=1&&board[r+1][c]==word.charAt(index)){roadMap[r+1][c]=1;flag|=backTraceWord(board,r+1,c,roadMap,word,index+1);if(flag==true) return true;roadMap[r+1][c]=0;}return flag;}}

解决方法二： 不用额外空间

public boolean exist(char[][] board, String word) {    char[] w = word.toCharArray();    for (int y=0; y<board.length; y++) {    for (int x=0; x<board[y].length; x++) {    if (exist(board, y, x, w, 0)) return true;    }    }    return false;}private boolean exist(char[][] board, int y, int x, char[] word, int i) {if (i == word.length) return true;if (y<0 || x<0 || y == board.length || x == board[y].length) return false;if (board[y][x] != word[i]) return false;board[y][x] ^= 256;boolean exist = exist(board, y, x+1, word, i+1)|| exist(board, y, x-1, word, i+1)|| exist(board, y+1, x, word, i+1)|| exist(board, y-1, x, word, i+1);board[y][x] ^= 256;return exist;}

参考：

【1】https://leetcode.com/