63. Unique Paths II

Medium (Trung bình) C++ 🔗 Xem trên LeetCode

📋 Đề Bài

You are given an m x n integer array grid. There is a robot initially located at the top-left corner (i.e., grid[0][0]). The robot tries to move to the bottom-right corner (i.e., grid[m - 1][n - 1]). The robot can only move either down or right at any point in time.

An obstacle and space are marked as 1 or 0 respectively in grid. A path that the robot takes cannot include any square that is an obstacle.

Return the number of possible unique paths that the robot can take to reach the bottom-right corner.

The testcases are generated so that the answer will be less than or equal to 2 * 109.

 

Example 1:

Input: obstacleGrid = [[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

Example 2:

Input: obstacleGrid = [[0,1],[0,0]]
Output: 1

 

Constraints:

  • m == obstacleGrid.length
  • n == obstacleGrid[i].length
  • 1 <= m, n <= 100
  • obstacleGrid[i][j] is 0 or 1.

🧠 Thuật Toán & Kỹ Thuật

Dynamic Programming (Quy hoạch động)Bit Manipulation (Thao tác bit)Matrix (Ma trận)
⏱️ Thời gian O(n×m)
💾 Không gian O(n×m)

💻 Lời Giải

C++ 0063-unique-paths-ii.cpp 
class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        // grid = [[0,0,0],[0,1,0],[0,0,0]]
        // grid[0] = [0,0,0]
        vector<vector<int>> dp(m, vector<int>(n, 0));
        for (int i = 0; i < n; ++i) {
            if (grid[0][i]) {
                break;
            }
            dp[0][i] = 1;
        }
        for (int i = 0; i < m; ++i) {
            if (grid[i][0]) {
                break;
            }
            dp[i][0] = 1;
        }
        for (int i = 1; i < m; ++i) {
            for (int j = 1; j < n; ++j) {
                if (grid[i][j] == 0) {
                    dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
                }
            }
        }
        return dp[m - 1][n - 1];
    }
};