problem
62. Unique Paths There is a robot on an m x n grid. The robot is initially located at the top-left corner (i.e., grid[0][0]). The robot tries to move to the bottom-right corner (i.e., grid[m - 1][n - 1]). The robot can only move either down or right at any point in time.
Given the two integers m and n, return the number of possible unique paths that the robot can take to reach the bottom-right corner.
The test cases are generated so that the answer will be less than or equal to 2 * 109.
Example 1:
Input: m = 3, n = 7
Output: 28
Example 2:
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 -> Down -> Down
2. Down -> Down -> Right
3. Down -> Right -> Down
approach permutation
class Solution {
public:
int uniquePaths(int m, int n) {
if(m == 1 || n == 1)
return 1;
m--, n--;
if(m < n) swap(m, n);
long res = 1;
for(int i = m+1, j=1; i <= m+n; i++, j++){
res *= i;
res /= j;
}
return (int)res;
}
};
approach 1 DP
class Solution {
public:
int uniquePaths(int m, int n) {
vector<vector<int>> dp = vector<vector<int>>(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];
}
};
approach 2 DP improve
class Solution {
public:
int uniquePaths(int m, int n) {
vector<int> cur(n, 1);
for (int i = 1; i < m; i++)
for (int j = 1; j < n; j++)
cur[j] += cur[j - 1];
return cur[n - 1];
}
};
explain: 1 1 1 1 1 1 2 3 4 5 1 3 6 10 15 14 10 20 35
|