动态规划 09
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https://leetcode-cn.com/problems/unique-paths/
动态规划
class Solution {
public int uniquePaths(int m, int n) {
int[][] dp = new int[n][m];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (i == 0 || j == 0) {
dp[i][j] = 1;
} else {
dp[i][j] = dp[i-1][j] + dp[i][j-1];
}
}
}
return dp[n-1][m-1];
}
}
数学方法(排列组合)
机器人无论如何都会往下走或者往右走,总共m + n - 2步,其中一定有m - 1步走右,剩余走下,或者n - 1走下,剩余走右,所以公式为:C (m+n-2) (n-1) 或 C (m+n-2) (m-1)
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