Dynamic Programming
Leetcode 62. Unique Paths
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).
How many possible unique paths are there?
class Solution {
public:
int uniquePaths(int m, int n) {
vector<vector<int>> dp(m+1, vector<int>(n+1, 0));
dp[1][0] = 1;
for (int i=1; i<m+1; i++)
for (int j=1; j<n+1; j++)
dp[i][j] = dp[i-1][j] + dp[i][j-1];
return dp[m][n];
}
};
Leetcode 63. Unique Paths II
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.
class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
if (obstacleGrid.empty()) return 0;
if (obstacleGrid[0][0] == 1) return 0;
int m = obstacleGrid.size();
int n = obstacleGrid[0].size();
for (int i=0; i<m; i++) {
for (int j=0; j<n; j++) {
if (i==0 && j==0) obstacleGrid[i][j] = 1;
else if (obstacleGrid[i][j]==0) {
obstacleGrid[i][j] = (i>=1? obstacleGrid[i-1][j]:0) + (j>=1 ? obstacleGrid[i][j-1]:0);
}
else
obstacleGrid[i][j] = 0;
}
}
return obstacleGrid[m-1][n-1];
}
};
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