leetcode Unique Paths II
2014-11-03 17:36
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Unique Paths II 原题地址:
https://oj.leetcode.com/problems/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
in the grid.
For example,
There is one obstacle in the middle of a 3x3 grid as illustrated below.
The total number of unique paths is
Note: m and n will be at most 100.
加了障碍。
没有到达障碍的路。
public class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int m = obstacleGrid.length;
int n = obstacleGrid[0].length;
if (obstacleGrid[m-1][n-1] == 1)
return 0;
int[][] paths = new int[m]
;
paths[m-1][n-1] = 1;
for (int i = m-2; i >= 0; i--) {
if (obstacleGrid[i][n-1] == 0)
paths[i][n-1] = paths[i+1][n-1];
else
paths[i][n-1] = 0;
}
for (int j = n-2; j >= 0; j--) {
if (obstacleGrid[m-1][j] == 0)
paths[m-1][j] = paths[m-1][j+1];
else
paths[m-1][j] = 0;
}
for (int i = m-2; i >= 0; i--)
for (int j = n-2; j >= 0; j--)
if (obstacleGrid[i][j] == 0)
paths[i][j] = paths[i+1][j] + paths[i][j+1];
else
paths[i][j] = 0;
return paths[0][0];
}
}
https://oj.leetcode.com/problems/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
1and
0respectively
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.
加了障碍。
没有到达障碍的路。
public class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int m = obstacleGrid.length;
int n = obstacleGrid[0].length;
if (obstacleGrid[m-1][n-1] == 1)
return 0;
int[][] paths = new int[m]
;
paths[m-1][n-1] = 1;
for (int i = m-2; i >= 0; i--) {
if (obstacleGrid[i][n-1] == 0)
paths[i][n-1] = paths[i+1][n-1];
else
paths[i][n-1] = 0;
}
for (int j = n-2; j >= 0; j--) {
if (obstacleGrid[m-1][j] == 0)
paths[m-1][j] = paths[m-1][j+1];
else
paths[m-1][j] = 0;
}
for (int i = m-2; i >= 0; i--)
for (int j = n-2; j >= 0; j--)
if (obstacleGrid[i][j] == 0)
paths[i][j] = paths[i+1][j] + paths[i][j+1];
else
paths[i][j] = 0;
return paths[0][0];
}
}
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