Leetcode (21)

Paths Unique

Topic

Analysis:

The robot has to go m+n-2 step, now for example analogy, there is a m+n-2 bit of binary number, now in which the M-bit fill 0, the rest of you fill 1, there is a C (m+n-2,m-1) kind of possible, if 0 means down, 1 means go right, so it is the same as the title.

Now consider the last step of the way, either to the right to reach the end, or down to the end, so

F (m,n) = f (m,n-1) +f (m-1,n);

The code is as follows (mainly considering big data):

1 classSolution {2  Public:3     intUniquepaths (intMintN) {4vector<vector<int>> V (M, vector<int> (n,1));5          for(intI=1; i<m; ++i) {6              for(intj=1; j<n; ++j) {7v[i][j]=v[i-1][j]+v[i][j-1];8             }9         }Ten         returnv[m-1][n-1]; One     } A};

Leetcode (21)