Leetcode——机器人的运动范围

Posted 2021-08-28 Yawn,

1. 题目

2. 题解

• 通过sum函数计算位数和，然后就是普通的dfs遍历

class Solution {
    int ans = 0;
    int[][] direction = {{1,0}, {-1,0}, {0,1}, {0,-1}};

    public int movingCount(int m, int n, int k) {
        boolean[][] vis = new boolean[m][n];
        dfs(m, n, vis, 0, 0, k);
        return ans;
    }

    //求数位和
    public int sums(int x){
        int s = 0;
        while(x != 0) {
            s += x % 10;
            x = x / 10;
        }
        return s;
    }

    public void dfs(int m, int n, boolean[][] vis, int x, int y, int k){
        if(sums(x) + sums(y) > k)
            return;

        //标记已经走过该位置，ans加一
        vis[x][y] = true;
        ans++;

        //四个方向遍历，找符合要求的坐标
        for(int[] dir : direction){
            int newX = x + dir[0];
            int newY = y + dir[1];
            if(newX >= 0 && newX < m && newY >= 0 && newY < n && !vis[newX][newY])
                dfs(m, n, vis, newX, newY, k);
        }
    }
}