【 LeetCode】79. Word Search

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本文介绍了一种使用回溯算法解决LeetCode上的单词搜索问题的方法。问题要求在一个二维字符网格中查找给定的单词，只能使用相邻的字符并且每个字符只能使用一次。通过递归回溯的方式检查所有可能路径来确定单词是否存在。

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问题描述

https://leetcode.com/problems/word-search/#/description

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.

算法

回溯算法解题

代码

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

        /**
         * 回溯算法进行搜索
         * @param board 字符表
         * @param visited 该字符是否已被访问
         * @param word 字符串
         * @param cur 当前字符,前面的0-(cur-1)已全部匹配，从(i,j)开始搜索
         * @param i 字符表的开始点行
         * @param j 字符表的开始点列
         * @return
         */
        private boolean exist(char[][] board, boolean[][] visited, String word, int cur, int i, int j) {
            if(cur == word.length()) {
                return true;
            }
            if(i<0||i>=board.length || j<0 || j >= board[0].length || visited[i][j]) {
                return false;
            }
            if(word.charAt(cur) != board[i][j]) {
                return false;
            }
            visited[i][j] = true;
            boolean b = exist(board, visited, word, cur+1, i+1, j) 
                    || exist(board, visited, word, cur+1, i, j+1)
                    || exist(board, visited, word, cur+1, i-1, j)
                    || exist(board, visited, word, cur+1, i, j-1);
            visited[i][j] = false;
            return b;
        }