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  • 【leetcode】Unique Paths

    Unique Paths

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

    Note: m and n will be at most 100.

     
    可以推到一下公式,对于m+1,n+1来说,只需要向右走m步,向下走n步,就能到达终点了,我们把向下和向右进行排列,所有的排列数就是结果
    因此result=(m+n)!/(m!*n!)
    直接求会超出范围,因此化简一下:
    化简一下(m+1)(m/2+1)……(m/n+1)
     
     
     1 class Solution {
     2 public:
     3    
     4     int uniquePaths(int m, int n) {
     5        
     6         int result=0;
     7         double tmp=1;
     8         m--;
     9         n--;
    10         for(int i=0;i<n;i++)
    11         {
    12             tmp*=(double)(m)/(i+1)+1;
    13         }
    14        
    15         result=round(tmp);
    16         return result;
    17     }
    18 };

     

     
    直接用动态规划
     
     1 class Solution {
     2 public:
     3    
     4     int uniquePaths(int m, int n) {
     5        
     6         if(m==0||n==0) return 0;
     7  
     8         int dp[101][101];
     9        
    10         dp[0][0]=1;
    11        
    12         for(int i=1;i<m;i++)
    13         {
    14             dp[i][0]=1;
    15         }
    16        
    17         for(int j=1;j<n;j++)
    18         {
    19             dp[0][j]=1;
    20         }
    21        
    22         for(int i=1;i<m;i++)
    23         {
    24             for(int j=1;j<n;j++)
    25             {
    26                 dp[i][j]=dp[i][j-1]+dp[i-1][j];
    27             }
    28         }
    29         return dp[m-1][n-1];
    30     }
    31 };
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  • 原文地址:https://www.cnblogs.com/reachteam/p/4202461.html
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