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).

How many possible unique paths are there?


Above is a 7 x 3 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

Example 1:

Input: m = 3, n = 2
Output: 3
Explanation:
From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
1. Right -> Right -> Down
2. Right -> Down -> Right
3. Down -> Right -> Right

Example 2:

Input: m = 7, n = 3
Output: 28

这道题让求所有不同的路径的个数,一开始还真把博主难住了,因为之前好像没有遇到过这类的问题,所以感觉好像有种无从下手的感觉。在网上找攻略之后才恍然大悟,原来这跟之前那道 Climbing Stairs 很类似,那道题是说可以每次能爬一格或两格,问到达顶部的所有不同爬法的个数。而这道题是每次可以向下走或者向右走,求到达最右下角的所有不同走法的个数。那么跟爬梯子问题一样,需要用动态规划 Dynamic Programming 来解,可以维护一个二维数组 dp,其中 dp[i][j] 表示到当前位置不同的走法的个数,然后可以得到状态转移方程为:  dp[i][j] = dp[i - 1][j] + dp[i][j - 1],这里为了节省空间,使用一维数组 dp,一行一行的刷新也可以,代码如下:

解法一:

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

这道题其实还有另一种很数学的解法,参见网友 Code Ganker 的博客,实际相当于机器人总共走了 m + n - 2步,其中 m - 1 步向右走,n - 1 步向下走,那么总共不同的方法个数就相当于在步数里面 m - 1 和 n - 1 中较小的那个数的取法,实际上是一道组合数的问题,写出代码如下:

解法二:

class Solution {
public:
int uniquePaths(int m, int n) {
double num = , denom = ;
int small = m > n ? n : m;
for (int i = ; i <= small - ; ++i) {
num *= m + n - - i;
denom *= i;
}
return (int)(num / denom);
}
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/62

类似题目:

Unique Paths II

Minimum Path Sum

Dungeon Game

参考资料:

https://leetcode.com/problems/unique-paths/

https://leetcode.com/problems/unique-paths/discuss/22981/My-AC-solution-using-formula

https://leetcode.com/problems/unique-paths/discuss/22954/0ms-5-lines-DP-Solution-in-C%2B%2B-with-Explanations

LeetCode All in One 题目讲解汇总(持续更新中...)

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