Leetcode Week10 Unique Paths II-优快云博客

本文探讨了在一个m x n网格中，机器人如何避开障碍物从起点到终点的路径规划问题。通过动态规划方法，计算出从任意一点到达终点的可行路径数，最终得出从起点到终点的总路径数。

Question

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 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

Answer

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[m-1][n-1] == 1) return 0;
        if (m == 1 && n == 1) return 1;
        vector<vector<int>> dp(m, vector<int>(n));
        if (n >= 2)
         dp[m-1][n-2] = (obstacleGrid[m-1][n-2] == 0) ? 1 : 0;
        if (m >= 2)
          dp[m-2][n-1] = (obstacleGrid[m-2][n-1] == 0) ? 1 : 0;
        for (int i = m-3; i >= 0; i--) {
            if (obstacleGrid[i][n-1] == 0) {
                dp[i][n-1] = dp[i+1][n-1];
            }
        }
          for (int i = n-3; i >= 0; i--) {
            if (obstacleGrid[m-1][i] == 0) {
                dp[m-1][i] = dp[m-1][i+1];
            }
        }
        for (int i = m-2; i >= 0; i--) {
            for (int j = n-2; j >= 0; j--) {
              if (obstacleGrid[i][j] == 0) {
                dp[i][j] = dp[i+1][j]+dp[i][j+1];
              }
            }
        }
        return dp[0][0];
    }
};