单词搜索

问题陈述

给定一个二维网格和一个单词，找出该单词是否存在于网格中。

单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。

示例：

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

给定 word = "ABCCED", 返回 true
给定 word = "SEE", 返回 true
给定 word = "ABCB", 返回 false

代码实现

DFS

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class Solution {
    //用于标记是否已经找到了解
    private boolean flag;

    public boolean exist(char[][] board, String word) {
        if(board == null || board.length == 0 || board[0].length == 0 ) {
            return false;
        }
        for(int i = 0; i < board.length; i++) {
            for(int j = 0; j < board[0].length; j++) {
                if(dfs(board, i, j, word, 0)) {
                    return true;
                }
            }
        }
        return false;
    }

    private boolean dfs(char[][] board, int i, int j, String word, int cur) {
        if(cur == word.length()) {
            flag = true;
            return true;
        }
        if(i < 0 || i >= board.length || j < 0 || j >= board[0].length 
                || board[i][j] != word.charAt(cur)) {
            return false;
        }
        //如果没有找到解，则继续DFS
        if(!flag) {
            char c = board[i][j];
            board[i][j] = '.';
            boolean ret1 = dfs(board, i + 1, j, word, cur + 1);
            boolean ret2 = dfs(board, i - 1, j, word, cur + 1);
            boolean ret3 = dfs(board, i, j + 1, word, cur + 1);
            boolean ret4 = dfs(board, i, j - 1, word, cur + 1);
            board[i][j] = c;
            return ret1 || ret2 || ret3 || ret4;
        }else {
            //找到解了，直接结束DFS并返回，这就是剪枝
            return true;
        }
    }
}


附：一.System.arraycopy（两数组合并）使用的基本定义
public static void arraycopy(Object src, int srcPos, Object dest, int destPos, int length)

src:源数组;

srcPos:源数组要复制的起始位置;

dest:目的数组;

destPos:目的数组放置的起始位置;

length:复制的长度.
————————————————

回溯

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public class Solution{
    private static final int[][] direction={{1,0},{-1,0},{0,1},{0,-1}};
    private int m;
    private int n;
    //主函数
    public boolean isExit(char[][] board,String word){
        if(word==null||word.length()==0){
            return false;
        }
        if(board==null||board.length==0||board[0].length==0){
            return false;
        }
        m=board.length;
        n=board[0].length;
        boolean[][] visited=new boolean[m][n];
        for(int r=0;r<m;r++){
            for(int c=0;c<n;c++){
                
            }
        }
    }
    //回溯
    private boolean backtracking(int curLen, int r, int c, boolean[][] visited, char[][] board, String word){
        //base case
        if(curLen==word.length()){
            return true;
        }
        if(r<0 || r>m || c<0 || c>n || visited[r][c] || board[r][c]!=word.charAt(curLen)){
            return false;
        }
        visited[r][c]=true;
        //选择
        for(int[] d:direction){
            if(backtracking(curLen+1,r+d[0],r+d[1],visited,board,word)){
                return true;
            }
        }
        //撤销
        visited[r][c]=false;
        return false;
    }
}