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<m;i++)dp[i][0]=1;
        for(int j=0;j<n;j++)dp[0][j]=1;
        for(int i=1;i<m;i++){
        for(int j=1;j<n;j++){
            if(i==0 && j==0)continue;
            dp[i][j] = dp[i-1][j]+dp[i][j-1];
        }
    }
        return (dp[m-1][n-1]);
    }
};