60、回溯-单词搜索

思路: 

这个是找到一种即可返回。那么可以双重便利，以任意一个位置i,j为起始点判断，开始上下左右遍历组成，每获取一个字符就匹配一次，匹配后的字符加个标记，下次不再匹配，相等则进行下一个元素获取。循环往复。代码如下：

class Solution {
       public boolean exist(char[][] board, String word) {
        if (board==null||board[0]==null||word==null||word.isEmpty()){
            return false;
        }
        char[] chs = word.toCharArray();
        for (int i = 0; i < board.length; i++) {
            char[] chars = board[i];
            for (int j = 0; j < chars.length; j++) {
                if ( process(board,i,j,chs,0)){
                   return true;
                }
            }
        }
        return false;
    }
 
    private boolean process(char[][] board, int i, int j, char[] chs, int index) {
        if (index==chs.length){
            return true;
        }
        if (i<0||i==board.length||j<0||j==board[0].length){
            return false;
        }
        if (board[i][j]!=chs[index]){
            return false;
        }
        char temp = board[i][j];
        board[i][j]=0;
        boolean ans=process(board,i+1,j,chs,index+1)
                ||process(board,i-1,j,chs,index+1)
                ||process(board,i,j+1,chs,index+1)
                ||process(board,i,j-1,chs,index+1);
        board[i][j]=temp;
        return ans;
    }
}