穷举vs暴搜vs深搜vs回溯vs剪枝系列一＞不同路径 III

整体思路：

代码设计：

代码呈现：

class Solution {
    int ret,step;
    int m,n;
    boolean[][] vis;

    public int uniquePathsIII(int[][] grid) {
        m = grid.length;
        n = grid[0].length;
        vis = new boolean[m][n];
        int bx = 0;
        int by = 0;
        for(int i = 0; i < m; i++)
            for(int j = 0; j < n; j++){
                if(grid[i][j] == 0) step++;
                else if(grid[i][j] == 1){
                    bx = i;
                    by = j;
                }
            }

        step += 2;    
        vis[bx][by] = true;
        dfs(grid,bx,by,1);
        return ret;    
    }

    int[] dx = {0,0,-1,1};
    int[] dy = {-1,1,0,0};
    public void dfs(int[][] grid,int i,int j, int count){

        if(grid[i][j] == 2) {
            if(count == step) ret++;
            return;
        }

        for(int k = 0; k < 4; k++){
            int x = i + dx[k];
            int y = j + dy[k];
            if(x >= 0 && x < m && y >= 0 && y < n && !vis[x][y] && grid[x][y] != -1){
                vis[i][j] = true;
                dfs(grid,x,y,count+1);
                vis[x][y] = false;
            }
        }
    }
}