62. Unique Paths
Problem:
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 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
思路:
利用DP算法。设二维数组dp[i][j]表示从起点到达点(i,j)的方法总数,则只能从点(i-1,j)或(i,j-1)到达点(i,j),故有dp[i][j]=dp[i-1][j]+dp[i][j-1],最后返回dp[m-1][n-1]即可。注意赋初值时,dp[i][0]=dp[0][j]=1,因为在边界上只有一种方法可以到达。
实际上用数学方法很好解释。从起点到中间总共要向右走n-1步,向下走m-1步,总共需要走m+n-2步,只需要计算出m+n-2步中m-1步的组合数即可,即\(C_{m+n-2}^{m-1}\)。
Solution:
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];
}
性能:
Runtime: 8 ms Memory Usage: 8.7 MB
知乎:littledy
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