给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中，返回 true ；否则，返回 false 。

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 1：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
输出：true

示例 2：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
输出：true

示例 3：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
输出：false

提示：

• m == board.length
• n = board[i].length
• 1 <= m, n <= 6
• 1 <= word.length <= 15
• board 和 word 仅由大小写英文字母组成

思路：原始思路是四个方向回溯，注意一旦找到word后续不用再进行搜索，立刻停止；但往往面试官会要求做优化，因此优化算法为统计word和board次数，如果一个字母在word中出现了x次，在board中出现了y次，且y<x，则不用再进行查找，直接返回false。

StringBuffer buffer=new StringBuffer();boolean [][]flag=new boolean[6][6];//使用used数组记录哪些已经使用过了，防止重复利用Map<Character,Integer> map=new HashMap<>();public boolean exist(char[][] board, String word) {// 优化：如果一个字母在word中出现了x次，在board中出现了y次，且y<x，则不用再进行查找，直接返回falsefor(int m=0;m<word.length();m++){if(!map.containsKey(word.charAt(m)))map.put(word.charAt(m),1);elsemap.put(word.charAt(m),map.get(word.charAt(m))+1);}for(int i=0;i<board.length;i++){for(int j=0;j<board[0].length;j++){if(map.containsKey(board[i][j]))map.put(board[i][j],map.get(board[i][j])-1);}}Set<Character> set=new HashSet<>();for(Character key :set){if(map.get(key)>0)return false;}for(int i=0;i<board.length;i++){for(int j=0;j<board[0].length;j++){if(board[i][j]==word.charAt(0)){buffer.append(board[i][j]);flag[i][j]=true;if(backTracking(board,word,i,j,1))return true;buffer.deleteCharAt(buffer.length()-1);flag[i][j]=false;}}}return false;}public boolean backTracking(char[][]board,String word,int i,int j,int index){if(buffer.length()==word.length())return true;// 四种情况搜索if(i+1<board.length&&board[i+1][j]==word.charAt(index)&&flag[i+1][j]==false){buffer.append(board[i+1][j]);flag[i+1][j]=true;if(backTracking(board,word,i+1,j,index+1))return true;buffer.deleteCharAt(buffer.length()-1);flag[i+1][j]=false;}if(i-1>=0&&board[i-1][j]==word.charAt(index)&&flag[i-1][j]==false){buffer.append(board[i-1][j]);flag[i-1][j]=true;if(backTracking(board,word,i-1,j,index+1))return true;buffer.deleteCharAt(buffer.length()-1);flag[i-1][j]=false;}if(j+1<board[0].length&&board[i][j+1]==word.charAt(index)&&flag[i][j+1]==false){buffer.append(board[i][j+1]);flag[i][j+1]=true;if(backTracking(board,word,i,j+1,index+1))return true;buffer.deleteCharAt(buffer.length()-1);flag[i][j+1]=false;}if(j-1>=0&&board[i][j-1]==word.charAt(index)&&flag[i][j-1]==false){buffer.append(board[i][j-1]);flag[i][j-1]=true;if(backTracking(board,word,i,j-1,index+1))return true;buffer.deleteCharAt(buffer.length()-1);flag[i][j-1]=false;}return false;}