62. Unique Paths

使用 领扣中国，来获得适合您的内容以及最佳的用户体验。
即刻前往   |   将我的账号同步到 LeetCode 中国
LeetCode
Explore
Problems
Mock 
Contest
Articles
Discuss
 Store 
 Premium
New Playground
lifeqiuzhi520
 874 67

62. Unique Paths
DescriptionHintsSubmissionsDiscussSolution
Pick One
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
Seen this question in a real interview before?  YesNo
Difficulty:Medium
Total Accepted:219.1K
Total Submissions:493.8K
Contributor:LeetCode
Subscribe to see which companies asked this question.

Related Topics 
ArrayDynamic Programming
Similar Questions 
Unique Paths IIMinimum Path SumDungeon Game
C++	





class Solution {
public:
     int uniquePaths(int m, int n) {
        if (m > n) return uniquePaths(n, m);
        vector<int> cur(m, 1);
        for (int j = 1; j < n; j++)
            for (int i = 1; i < m; i++)
                cur[i] += cur[i - 1]; 
        return cur[m - 1];
    }
};
1
class Solution {
2
public:
3
     int uniquePaths(int m, int n) {
4
        if (m > n) return uniquePaths(n, m);
5
        vector<int> cur(m, 1);
6
        for (int j = 1; j < n; j++)
7
            for (int i = 1; i < m; i++)
8
                cur[i] += cur[i - 1]; 
9
        return cur[m - 1];
10
    }
11
};
  Custom Testcase( Contribute  )
 Run Code Submit Solution
Submission Result: Accepted More Details 
Next challenges: Unique Paths IIMinimum Path SumDungeon Game
Share your acceptance!
Notes
|||

Type here...(Markdown is enabled)
​
Copyright © 2018 LeetCode Contact Us  |  Jobs  |  Students  |  Frequently Asked Questions  |  Terms of Service  |  Privacy Policy      
译