leetcode 62|63. Unique Paths 1|2
阿新 • • 發佈:2019-02-18
62. Unique Paths
標準動態規劃,當前步只能從上面和左邊走過來。
class Solution {
public:
int uniquePaths(int m, int n)
{
vector<vector<int>> ret(n, vector<int> (m, 0));
for (int i = 0; i < m; i++)
ret[0][i] = 1;
for (int i = 0; i < n; i++)
ret[i][0] = 1;
for (int i = 1; i < n; i++)
{
for(int j = 1; j < m; j++)
{
ret[i][j] = ret[i-1][j] + ret[i][j-1];
}
}
return ret[n-1][m-1];
}
};
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
.
Note: m and n will be at most 100.
還是動態規劃,加一個條件而已class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid)
{
int row = obstacleGrid.size();
int col = obstacleGrid[0].size();
if (row > 0 && col > 0 && obstacleGrid[0][0] == 1)
return 0;
int flag = 1;
for (int i = 0; i < row; i++)
{
if (flag == 1 && obstacleGrid[i][0] == 0)
obstacleGrid[i][0] = 1;
else
{
flag = 0;
obstacleGrid[i][0] = 0;
}
}
flag = 1;
for(int i = 1; i < col; i++)
{
if (flag == 1 && obstacleGrid[0][i] == 0)
obstacleGrid[0][i] = 1;
else
{
flag = 0;
obstacleGrid[0][i] = 0;
}
}
for(int i = 1; i < row; i++)
for(int j = 1; j < col; j++)
{
if (obstacleGrid[i][j] == 1)
obstacleGrid[i][j] = 0;
else
obstacleGrid[i][j] = obstacleGrid[i-1][j] + obstacleGrid[i][j-1];
}
return obstacleGrid[row-1][col-1];
}
};