【劍指offer】65. 矩陣中的路徑

問題描述

思路

DFS經典問題。沒啥好說的。說點技巧把。 方向數組的應用，剪枝的應用。

方法一

import java.util.*;
public class Solution {
    private int rows, cols;
    public boolean hasPath(char[] matrix, int rows, int cols, char[] str) {
        if(matrix == null || matrix.length == 0 || str == null || str.length == 0) return false;
        this.rows = rows; this.cols = cols;
        for(int row = 0; row < rows; row++){
            for(int col = 0; col < cols; col++){
                if(matrix[getIndex(row,col)] == str[0]){
                    if(logic(matrix, row, col, str, 0, new boolean[matrix.length])) return true;
                }
            }
        }
        return false;
    }
    public boolean logic(char[] matrix, int row, int col, char[] str, int strIndex, boolean[] isVisited){
        if(isVisited[getIndex(row,col)] || matrix[getIndex(row,col)] != str[strIndex]) return false;
        // 能走到這裏, 證明肯定是相等的
        if(strIndex == str.length-1) return true;
        isVisited[getIndex(row,col)] = true;
        int[] direcX = new int[]{1,0,-1,0};
        int[] direcY = new int[]{0,1,0,-1};
        for(int step = 0; step < 4; step++){
            int curX = row + direcX[step], curY = col + direcY[step];
            if(curX<0 || curX>=rows || curY<0 || curY>=cols || isVisited[getIndex(curX,curY)]) continue;
            // 剪枝操作
            if(logic(matrix, curX, curY, str, strIndex+1, isVisited)) return true;
            isVisited[getIndex(curX,curY)] = false;
        }
        return false;
    }
    private int getIndex(int row, int col){
        return row * cols + col;
    }
}