LeetCode 62 Unique Paths DP
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 * 10^9.
Solution
\(dp[i][j]\) 表示到位置 \(i,j\) 的方案数，很明显的转移方程：
\[dp[i][j] = dp[i-1][j]+dp[i][j-1] \]对于初始化，由于只能右或者下，所以边界的方案数都为 \(1\):
\[dp[i][0] = dp[0][j] = 1 \]点击查看代码
class Solution {
public:
int uniquePaths(int m, int n) {
long long dp[105][105];
if(m==1&&n==1)return 1;
for(int i=0;i