Leetcode-63. Unique Paths II

前言:为了后续的实习面试,开始疯狂刷题,非常欢迎志同道合的朋友一起交流。因为时间比较紧张,目前的规划是先过一遍,写出能想到的最优算法,第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN,mcf171专栏。

博客链接:mcf171的博客

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Follow up for "Unique Paths":

Now consider if some obstacles are added to the grids. How many unique paths would there be?

An obstacle and empty space is marked as 1 and 0 respectively in the grid.

For example,

There is one obstacle in the middle of a 3x3 grid as illustrated below.

[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]

The total number of unique paths is 2.

Note: m and n will be at most 100.

这个题目第一反应是能不能用62的数学方法做,稍微考虑了下太麻烦了,还是用递归做算了。构建一个类似树的结构,进行深度优先遍历。对于访问过的节点保存其后续的路径数量,从而避免再次访问。 Your runtime beats 14.02% of java submissions.

public class Solution {
    public class TreeNode{
        int down;
        int right;
        TreeNode(int down,int right){
            this.down = down;
            this.right = right;
        }
    }

    public int uniquePathsWithObstacles(int[][] obstacleGrid) {

        int m = obstacleGrid.length;
        if(m == 0) return 0;
        int n = obstacleGrid[0].length;
        if(obstacleGrid[0][0] == 1) return 0;
        TreeNode start = new TreeNode( m - 1, n - 1);
        int results = goDeep(start,0,0,obstacleGrid);

        return results;
    }

    public int goDeep(TreeNode node, int i, int j, int[][]nums){
        if(node.down ==0 && node.right ==0)
            if(nums[i][j] != 1)return 1;
            else return 0;
        int downPath = 0,rightPath = 0;
        if(node.down > 0 && nums[i+1][j] != 1) {
            if(nums[i+1][j] == 0) {
                downPath = goDeep(new TreeNode(node.down - 1, node.right), i + 1, j, nums);
                nums[i + 1][j] = downPath == 0 ? nums[i+1][j] : -downPath;
            }else downPath = -nums[i+1][j];

        }
        if(node.right > 0 && nums[i][j+1] !=1) {
            if(nums[i][j+1] == 0) {
                rightPath = goDeep(new TreeNode(node.down, node.right - 1), i, j + 1, nums);
                nums[i][j + 1] = rightPath == 0 ? nums[i][j+1] : -rightPath;
            }else rightPath = -nums[i][j+1];
        }

        int result = downPath + rightPath;
        if (result ==0)
            nums[i][j] = 1;
        return result;

    }
}







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