[LeetCode] (medium) 63. Unique Paths II

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本文探讨了一个机器人在含有障碍物的网格中寻找从起点到终点的所有可能路径的问题。通过结合深度优先搜索（DFS）和动态规划（DP），提出了一种有效的解决方案，避免了传统DP方法在某些情况下可能出现的整数溢出问题。

https://leetcode.com/problems/unique-paths-ii/submissions/

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).

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 1and 0 respectively in the grid.

Note: m and n will be at most 100.

Example 1:
Input:
[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]
Output: 2
Explanation:
There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:
1. Right -> Right -> Down -> Down
2. Down -> Down -> Right -> Right

直接用dp会在某些中间个点中发生int溢出，这是因为这些中间节点不存在到达终点的道路所以它的值本身是无关紧要的。我又不想把dp数组设置成long long类型，所以使用了dfs+dp的方法（我忘记专有名词了）反向查找，只关心那些可以到达终点的个点的中间值。

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        if(obstacleGrid.size() == 0) return 0;
        
        int row = obstacleGrid.size();
        int col = obstacleGrid[0].size();
        
        if(obstacleGrid[0][0] == 1 || obstacleGrid[row-1][col-1] == 1) return 0;    //这个样例真的太傻比了
        
        vector<vector<int>> result(row, vector<int>(col, 0));
        vector<vector<bool>> visited(row, vector<bool>(col, false));
        
        
        result[0][0] = 1;
        visited[0][0] = true;
        
        return dfs(result, visited, obstacleGrid, row-1, col-1);
        
    }
    
    int dfs(vector<vector<int>> &result, vector<vector<bool>> &visited, vector<vector<int>>& obstacleGrid, int r, int c){
        if(r < 0 || c < 0){
            return 0;
        }else if(visited[r][c]){
            return result[r][c];
        }else if(obstacleGrid[r][c] == 1){
            return 0;
        }else{
            result[r][c] = dfs(result, visited, obstacleGrid, r-1, c) + dfs(result, visited, obstacleGrid, r, c-1);
            visited[r][c] = true;
            return result[r][c];
        }
    }
};