79. Word Search

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.

思路：这道题和剑指Offer上的面试题66：矩阵中的路径是相似的题目，都是采用回溯法
用递归实现回溯好写一点，感觉是O(n2)级别的时间复杂度呢

Java代码如下：

public class Solution {    public boolean exist(char[][] board, String word) {        int rows = board.length;        int cols = board[0].length;        boolean[][] visited = new boolean[rows][cols];        int len = 0;        for(int i = 0; i < rows; i++) {            for(int j = 0; j < cols; j++) {                if(existCore(board, i, j, rows, cols, word, len, visited)) {                    return true;                }            }        }        return false;    }    // 递归方法    public boolean existCore(char[][] board, int i, int j, int rows, int cols,                             String word, int len, boolean[][] visited) {        if(len == word.length()) {            return true;        }        boolean result = false;        if(i >= 0 && i < rows && j >=0 && j < cols &&          board[i][j] == word.charAt(len) && !visited[i][j]) {            visited[i][j] = true;            len++;            result = existCore(board, i-1, j, rows, cols, word, len, visited) ||                     existCore(board, i, j-1, rows, cols, word, len, visited) ||                     existCore(board, i+1, j, rows, cols, word, len, visited) ||                     existCore(board, i, j+1, rows, cols, word, len, visited);            if(!result) {                len--;                visited[i][j] = false;            }        }        return result;    }}