康瑞部落

Unique Paths II

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.

解题思路：

这道题的题意是从左上角到右下角走，一次只能走一步，且方向只能向下或向右。1是障碍物，不能行走。一共有多少种走法。

方法1：回溯法。代码如下。但是会产生超时问题。

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m = obstacleGrid.size();
        if(m<=0){
            return 0;
        }
        int n = obstacleGrid[0].size();
        if(n<=0){
            return 0;
        }
        if(obstacleGrid[0][0]==1 || obstacleGrid[m-1][n-1]==1){
            return 0;
        }
        
        int result = 0;
        
        helper(obstacleGrid, 0, 0, m, n, result);
        
        return result;
    }
    
    void helper(vector<vector<int>>& obstacleGrid, int i, int j, int m, int n, int& result){
        if(obstacleGrid[i][j]==1){
            return;
        }
        if(i==m-1&&j==n-1){
            result++;
            return;
        }
        if(i<m-1){
            helper(obstacleGrid, i+1, j, m, n, result);
        }
        if(j<n-1){
            helper(obstacleGrid, i, j+1, m, n, result);
        }
    }
};

方法2：动态规划法。设d[i][j]表示从0,0到i,j共有多少条路径。则有:

d[i][j]=0, 当matrix[i][j]=1;

d[i][j]=d[i-1][j] + d[i][j-1]; 其他。

注意边界条件的限制。

代码如下：

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m = obstacleGrid.size();
        if(m<=0){
            return 0;
        }
        int n = obstacleGrid[0].size();
        if(n<=0){
            return 0;
        }
        if(obstacleGrid[0][0]==1 || obstacleGrid[m-1][n-1]==1){
            return 0;
        }
        vector<vector<int>> d(m, vector<int>(n, 0));
        d[0][0] = 1;
        for(int i=0; i<m; i++){
            for(int j=0; j<n; j++){
                if(obstacleGrid[i][j]==1){
                    d[i][j]=0;
                }else{
                    if(i>0){
                        d[i][j] += d[i-1][j];
                    }
                    if(j>0){
                        d[i][j] += d[i][j-1];
                    }
                }
            }
        }
        return d[m-1][n-1];
    }
};