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).
Now consider if some obstacles are added to the grids. How many unique paths would there be?
An obstacle and empty space is marked as 1 and 0 respectively in the grid.
Note: m and n will be at most 100.
Example 1:
Input: [ [0,0,0], [0,1,0], [0,0,0] ] Output: 2 Explanation: There is one obstacle in the middle of the 3x3 grid above. There are two ways to reach the bottom-right corner: 1. Right -> Right -> Down -> Down 2. Down -> Down -> Right -> Right这个题状态于62像是，但是有1的地方是障碍点不能通过class Solution {public:int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {int m=obstacleGrid.size();int n=obstacleGrid[0].size();vector<vector<int> >dp (m,vector<int>(n));for(int i=0;i<m;i++){if(obstacleGrid[i][0]==1)break;elsedp[i][0]=1;}for(int j=0;j<n;j++){if(obstacleGrid[0][j]==1)break;elsedp[0][j]=1;}for(int i=1;i<m;i++)for(int j=1;j<n;j++){if(obstacleGrid[i][j]==1)dp[i][j]=0;elsedp[i][j]=dp[i-1][j]+dp[i][j-1];}return dp[m-1][n-1];}};但是这里会报错Line 27: Char 23: runtime error: signed integer overflow: 1053165744 + 1579748616 cannot be represented in type 'int' (solution.cpp)int类型精度不够，我们把dp容器的int 换成long就ac了
class Solution {public: int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) { int m=obstacleGrid.size(); int n=obstacleGrid[0].size(); vector<vector<long> >dp (m,vector<long>(n)); for(int i=0;i<m;i++) { if(obstacleGrid[i][0]==1) break; else dp[i][0]=1; } for(int j=0;j<n;j++) { if(obstacleGrid[0][j]==1) break; else dp[0][j]=1; } for(int i=1;i<m;i++) for(int j=1;j<n;j++) { if(obstacleGrid[i][j]==1) dp[i][j]=0; else dp[i][j]=dp[i-1][j]+dp[i][j-1]; } return dp[m-1][n-1]; }};
转载于:https://www.cnblogs.com/flyljz/p/10981232.html
