Skip to content

Latest commit

 

History

History
175 lines (140 loc) · 4.97 KB

File metadata and controls

175 lines (140 loc) · 4.97 KB

English Version

题目描述

一个机器人位于一个 m x n 网格的左上角 (起始点在下图中标记为“Start” )。

机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角(在下图中标记为“Finish”)。

现在考虑网格中有障碍物。那么从左上角到右下角将会有多少条不同的路径?

网格中的障碍物和空位置分别用 10 来表示。

 

示例 1:

输入:obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]]
输出:2
解释:
3x3 网格的正中间有一个障碍物。
从左上角到右下角一共有 2 条不同的路径:
1. 向右 -> 向右 -> 向下 -> 向下
2. 向下 -> 向下 -> 向右 -> 向右

示例 2:

输入:obstacleGrid = [[0,1],[0,0]]
输出:1

 

提示:

  • m == obstacleGrid.length
  • n == obstacleGrid[i].length
  • 1 <= m, n <= 100
  • obstacleGrid[i][j]01

解法

动态规划。

假设 dp[i][j] 表示到达网格 (i,j) 的路径数,先初始化 dp 第一列和第一行的所有值,然后判断。

  • obstacleGrid[i][j] == 1,说明路径数为 0,dp[i][j] = 0
  • obstacleGrid[i][j] == 0,则 dp[i][j] = dp[i - 1][j] + dp[i][j - 1]

最后返回 dp[m - 1][n - 1] 即可。

Python3

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

Java

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length, n = obstacleGrid[0].length;
        int[][] dp = new int[m][n];
        for (int i = 0; i < m && obstacleGrid[i][0] == 0; ++i) {
            dp[i][0] = 1;
        }
        for (int j = 0; j < n && obstacleGrid[0][j] == 0; ++j) {
            dp[0][j] = 1;
        }
        for (int i = 1; i < m; ++i) {
            for (int j = 1; j < n; ++j) {
                if (obstacleGrid[i][j] == 0) {
                    dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
                }
            }
        }
        return dp[m - 1][n - 1];
    }
}

C++

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

Go

func uniquePathsWithObstacles(obstacleGrid [][]int) int {
	m, n := len(obstacleGrid), len(obstacleGrid[0])
	dp := make([][]int, m)
	for i := 0; i < m; i++ {
		dp[i] = make([]int, n)
	}
	for i := 0; i < m && obstacleGrid[i][0] == 0; i++ {
		dp[i][0] = 1
	}
	for j := 0; j < n && obstacleGrid[0][j] == 0; j++ {
		dp[0][j] = 1
	}
	for i := 1; i < m; i++ {
		for j := 1; j < n; j++ {
			if obstacleGrid[i][j] == 0 {
				dp[i][j] = dp[i-1][j] + dp[i][j-1]
			}
		}
	}
	return dp[m-1][n-1]
}

...