LeetCode 63 Unique Paths II DP

You are given an m x n integer array grid. There is a robot 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.

An obstacle and space are marked as 1 or 0 respectively in grid. A path that the robot takes cannot include any square that is an obstacle.

Return the number of possible unique paths that the robot can take to reach the bottom-right corner.

The testcases are generated so that the answer will be less than or equal to 2 * 109.

Solution

显然设 \(dp[i][j]\) 表示到 \(grid[i][j]\) 的方案数，转移方程：

\[dp[i][j] = dp[i-1][j]+dp[i][j-1] \]

点击查看代码

class Solution {
private:
    int dp[103][103];
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int n = obstacleGrid.size(), m = obstacleGrid[0].size();
        if(n==1 && m==1){
            if(obstacleGrid[0][0]==1) return 0;
            else return 1;
        }
        if(obstacleGrid[0][0]==1)return 0;
        else dp[0][0]=1;
        for(int i=1;i<n;i++){
            if(obstacleGrid[i][0]==0 && dp[i-1][0])dp[i][0]=1;
            else dp[i][0]=0;
        }
        for(int j=1;j<m;j++){
            if(obstacleGrid[0][j]==0 && dp[0][j-1])dp[0][j]=1;
            else dp[0][j]=0;
        }
        for(int i=1;i<n;i++){
            for(int j=1;j<m;j++){
                if(obstacleGrid[i][j]==0){
                    dp[i][j]=dp[i-1][j]+dp[i][j-1];
                }
                else dp[i][j]=0;
            }
        }
        return dp[n-1][m-1];
    }
};