62. Unique Paths
Description
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?
Analysis
这道题的意思就是问有多少种方法可以从左上角到右下角。
限制就是只能向下或者向右走。
所以我们不难看出,其实当我们想求到达m,n时的位置的不同方法时,
其实它就等于到达m-1,n的方法数和到达m,n-1的方法数之和。
所以我们可以利用这个关系来求到达m,n的方法数。
注意,我们到达m,1或者到达n,1的方法数都是1。
Code
class Solution {
public:
int uniquePaths(int m, int n) {
int res [m][n];
res[0][0] = 0;
//res[0][1] = 1;
//if(m!=1)res[1][0] = 1;
for(int i = 1 ; i < n ;++i){
res[0][i] = 1;
}
for(int i = 1 ; i < m ;++i){
res[i][0] = 1;
}
for(int i =1;i<m;++i){
for(int j = 1;j<n;++j){
res[i][j] = res[i-1][j] + res[i][j-1];
}
}
return res[m-1][n-1];
}
};