12 face questions backtracking

 

/**
 * @ClassName Solution
 * @Author wangyudi
 * @Date 2019/7/14 16:13
 * @Version 1.0
 * @Description
 */

class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;
    }

}

public class Solution {
    public boolean findPath(char[][] matrix, String str) {
        //INPUT with invalid Deal 
        IF (STR == null || Matrix == null ) return  to false ;
         // The following default input is a rectangular two-dimensional array of valid 
        int rows = matrix.length;
         int cols = Matrix [0 ] .length;
         Boolean [] = visited new new  Boolean [rows cols *]; // stored the array element corresponding to whether the information has been accessed 
        char [] = charStr str.toCharArray ();
         int pathIndex = 0; // record the need to find the starting end of the string 
        Boolean Result = to false ;
         for ( int row = 0; row < rows; row++) {
            for (int col = 0; col < cols; col++) {
                if (findPathCore(matrix, charStr, pathIndex, row, col, rows, cols, visited)) {
                    result = true;
                    break;
                }
            }
        }
        return result;
    }

    /**
     * 查找路径的迭代函数
     * @param matrix
     * @param charStr
     * @param pathIndex
     * @param row
     * @Param COL 
     * @param rows 
     * @param cols 
     * @param visited 
     * @return 
     * / 
    Private  Boolean findPathCore ( char [] [] Matrix, char [] charStr, int pathIndex, int Row, int COL, int rows, int cols, Boolean [] visited) {
         // if judged to have been found string 1 
        iF (pathIndex == charStr.length) return  to true ;
         Boolean= Result to false ;
         // 2-dimensional array of characters is determined whether the corresponding character string 
        IF (Row> Row = 0 && <&& rows 
                COL > = 0 && COL <cols && 
                charStr [pathIndex] == Matrix [Row] [COL] && 
                visited [Row * cols + COL] == to false ) {
             // corresponds
             // find next substring, and modify the access information 
            pathIndex ++ ; 
            visited [Row * cols + COL] = to true ;
             // from the four-direction position of the element to find the next string 
            result = findPathCore (matrix, charStr, pathIndex, row + 1, col, rows, cols,visited) ||
                    findPathCore (Matrix, charStr, pathIndex, Row -. 1, COL, rows, cols, visited) || 
                    findPathCore (Matrix, charStr, pathIndex, Row, COL +. 1, rows, cols, visited) || 
                    findPathCore (Matrix, charStr, pathIndex, Row, COL -. 1 , rows, cols, visited);
             // four directions are no corresponding substring is found, it indicates that the location is not correct
             // modified access information, and then return to a position 
            IF (! Result) { 
                visited [Row * cols + COL] = to false ; 
            } 
        } 
        return Result; 
    } 

// the Test Output INPUT and 
    public static void main(String[] args) {
        Solution solution = new Solution();
        char[] char1 = {'a', 'b', 't', 'g'};
        char[] char2 = {'c', 'f', 'c', 's'};
        char[] char3 = {'j', 'd', 'e', 'h'};
        char[] char4 = {'a', 'b', 't', 'g'};
        char[][] chars = {char1, char2, char3, char4};
        String str = "bfcc";
        System.out.println(solution.findPath(chars, str));
    }

}