【 LeetCode】79. Word Search

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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;
        }