【LeetCode】063. Unique Paths II
题目:
Follow up for "Unique Paths":
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.
For example,
There is one obstacle in the middle of a 3x3 grid as illustrated below.
- [
- [0,0,0],
- [0,1,0],
- [0,0,0]
- ]
The total number of unique paths is 2
.
Note: m and n will be at most 100.
题解:
Solution 1 ()(未优化)
- class Solution {
- public:
- int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
- if(obstacleGrid.empty() || obstacleGrid[].empty()) return ;
- int m = obstacleGrid.size(), n = obstacleGrid[].size();
- vector<int> dp(n,);
- dp[] = ;
- for(int i=; i<m; ++i) {
- for(int j=; j<n; ++j) {
- if(obstacleGrid[i][j] == )
- dp[j] = dp[j-];
- else dp[j] += dp[j-];
- }
- }
- return dp[n-];
- }
- };
Solution 2 ()
- class Solution {
- public:
- int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
- if(obstacleGrid.empty() || obstacleGrid[].empty()) return ;
- int m = obstacleGrid.size(), n = obstacleGrid[].size();
- vector<int> dp(n,);
- dp[] = ;
- for(int i=; i<m; ++i) {
- for(int j=; j<n; ++j) {
- if(obstacleGrid[i][j] == )
- dp[j] = ;
- else {
- if(j > )
- dp[j] += dp[j-];
- }
- }
- }
- return dp[n-];
- }
- };
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