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62. Unique Paths

时间:2019-10-26 14:31:32      阅读:75      评论:0      收藏:0      [点我收藏+]

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?

技术分享图片
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



二刷,本来也算是启蒙题目,差点没做出来哈哈哈哈. 刷题看来得一直刷,不然很容易丧失这种思考模式
class Solution {
    
public:
    int uniquePaths(int m, int n) {
        vector<vector<int>> dp(n+1,vector<int>(m+1,0));            dp[1][1]=1;
        for(int i=1;i<=n;++i)
            for(int j=1;j<=m;++j)
            {
                if(i>1)dp[i][j]+=dp[i-1][j];
                if(j>1)dp[i][j]+=dp[i][j-1];
            }
        return dp[n][m];
    }
};

 

讨论区更加subtle的解法..

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];
    }
};

 

62. Unique Paths

原文:https://www.cnblogs.com/lychnis/p/11742873.html

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