0063. Unique Paths II.md

• Unique Paths II
• 难度：Medium｜ 中等
• 相关知识点：数组 动态规划
• 题目链接：https://leetcode-cn.com/problems/unique-paths-ii/
• 官方题解：https://leetcode-cn.com/problems/unique-paths-ii/solution/

Problem Statement:

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 space are marked as 1 and 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

Solution`:

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        m = len(obstacleGrid)
        n = len(obstacleGrid[0])

        dp = [0 for i in range(m)]
        dp[0] = 1 if (obstacleGrid[0][0] ==0) else 0

        for k in range(n):
            for j in range(m):
                if obstacleGrid[j][k] == 1:
                    dp[j] = 0
                else:
                    if j >= 1:
                        dp[j] += dp[j-1]
        return dp[-1]

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

Solution2:

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        m = len(obstacleGrid)
        n = len(obstacleGrid[0])
        
        dp = [[0] * (n + 1) for _ in range(m + 1)]
        dp[0][1] = 1  # Can also set dp[1][0] = 1
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if obstacleGrid[i - 1][j - 1] == 0:
                    dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
        return dp[-1][-1]