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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” )。

问总共有多少条不同的路径?

 

示例 1:

输入:m = 3, n = 7
输出:28

示例 2:

输入:m = 3, n = 2
输出:3
解释:
从左上角开始,总共有 3 条路径可以到达右下角。
1. 向右 -> 向下 -> 向下
2. 向下 -> 向下 -> 向右
3. 向下 -> 向右 -> 向下

示例 3:

输入:m = 7, n = 3
输出:28

示例 4:

输入:m = 3, n = 3
输出:6

 

提示:

  • 1 <= m, n <= 100
  • 题目数据保证答案小于等于 2 * 109

解法

动态规划。

假设 dp[i][j] 表示到达网格 (i,j) 的路径数,则 dp[i][j] = dp[i - 1][j] + dp[i][j - 1]

Python3

class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        dp = [[1] * n for _ in range(m)]
        for i in range(1, m):
            for j in range(1, n):
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
        return dp[-1][-1]

Java

class Solution {
    public int uniquePaths(int m, int n) {
        int[][] dp = new int[m][n];
        for (int i = 0; i < m; ++i) {
            Arrays.fill(dp[i], 1);
        }
        for (int i = 1; i < m; ++i) {
            for (int j = 1; j < n; ++j) {
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
            }
        }
        return dp[m - 1][n - 1];
    }
}

TypeScript

function uniquePaths(m: number, n: number): number {
    let dp = Array.from({ length: m }, v => new Array(n).fill(1));
    for (let i = 1; i < m; ++i) {
        for (let j = 1; j < n; ++j) {
            dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
        }
    }
    return dp[m - 1][n - 1];
}

C++

class Solution {
public:
    int uniquePaths(int m, int n) {
        vector<vector<int>> dp(m, vector<int>(n, 1));
        for (int i = 1; i < m; ++i)
        {
            for (int j = 1; j < n; ++j)
            {
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
            }
        }
        return dp[m - 1][n - 1];
    }
};

Go

func uniquePaths(m int, n int) int {
	dp := make([][]int, m)
	for i := 0; i < m; i++ {
		dp[i] = make([]int, n)
	}
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			if i == 0 || j == 0 {
				dp[i][j] = 1
			} else {
				dp[i][j] = dp[i-1][j] + dp[i][j-1]
			}
		}
	}
	return dp[m-1][n-1]
}

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