[LeetCode] 63.Unique Paths II
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
62. Unique Paths 的拓展， 有以下不同：
1. 当(i, j)有障碍时dp[i][j] = 0
2. dp[0][j]和dp[i][0]未必为1.
dp[0][j] = obstacleGrid[0][j] ? 0 : dp[0][j-1]
dp[i][0] = obstacleGrid[i][0] ? 0 : dp[i-1][0]
3. 当obstacleGrid [0][0] = 1时，return 0
解法：DP, 建立二维数组，dp[i][j]表示在某一位置能到达的不同路径数量，dp[i][j] = dp[i-1][j] + dp[i][j-1] ，如果某一位置grid[i][j]=1说明为障碍物，那么dp[i][j] = 0，还要注意i, j为0的情况。
Java:
class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int m = obstacleGrid.length;
int n = obstacleGrid[0].length;
if (m == 0 || n == 0) {
return 0;
}
if (obstacleGrid[0][0] == 1 || obstacleGrid[m-1][n-1] == 1) {
return 0;
}
int[][] dp = new int[m][n];
dp[0][0] = 1;
for(int i = 1; i < n; i++){
if(obstacleGrid[0][i] == 1) {
dp[0][i] = 0;
}
else {
dp[0][i] = dp[0][i-1];
}
}
for(int i = 1; i < m; i++){
if(obstacleGrid[i][0] == 1) {
dp[i][0] = 0;
}
else {
dp[i][0] = dp[i-1][0];
}
}
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][j-1] + dp[i-1][j];
}
}
}
return dp[m-1][n-1];
}
}
Java:
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int width = obstacleGrid[0].length;
int[] dp = new int[width];
dp[0] = 1;
for (int[] row : obstacleGrid) {
for (int j = 0; j < width; j++) {
if (row[j] == 1)
dp[j] = 0;
else if (j > 0)
dp[j] += dp[j - 1];
}
}
return dp[width - 1];
}
Python:
class Solution:
def uniquePathsWithObstacles(self, obstacleGrid):
mp = obstacleGrid
for i in range(len(mp)):
for j in range(len(mp[i])):
if i == 0 and j == 0:
mp[i][j] = 1 - mp[i][j]
elif i == 0:
if mp[i][j] == 1:
mp[i][j] = 0
else:
mp[i][j] = mp[i][j - 1]
elif j == 0:
if mp[i][j] == 1:
mp[i][j] = 0
else:
mp[i][j] = mp[i - 1][j]
else:
if mp[i][j] == 1:
mp[i][j] = 0
else:
mp[i][j] = mp[i - 1][j] + mp[i][j - 1]
if mp[-1][-1] > 2147483647:
return -1
else:
return mp[-1][-1]
Python: wo
class Solution(object):
def uniquePathsWithObstacles(self, obstacleGrid):
"""
:type obstacleGrid: List[List[int]]
:rtype: int
"""
m, n = len(obstacleGrid), len(obstacleGrid[0])
dp = [[0] * n for i in xrange(m)]
for i in xrange(m):
for j in xrange(n):
if obstacleGrid[i][j] == 1:
dp[i][j] == 0
else:
if i == 0 and j == 0: # 写错成了 or
dp[i][j] = 1
elif i == 0:
dp[i][j] = dp[i][j-1]
elif j == 0:
dp[i][j] = dp[i-1][j]
else:
dp[i][j] = dp[i-1][j] + dp[i][j-1]
return dp[-1][-1]
Python: wo
class Solution(object):
def uniquePathsWithObstacles(self, obstacleGrid):
"""
:type obstacleGrid: List[List[int]]
:rtype: int
"""
m, n = len(obstacleGrid), len(obstacleGrid[0])
dp = [[0] * n for i in xrange(m)]
for i in xrange(m):
for j in xrange(n):
if obstacleGrid[i][j] != 1:
if i == 0 and j == 0:
dp[i][j] = 1
elif i == 0:
dp[i][j] = dp[i][j-1]
elif j == 0:
dp[i][j] = dp[i-1][j]
else:
dp[i][j] = dp[i-1][j] + dp[i][j-1]
return dp[-1][-1]
C++:
class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
int m = obstacleGrid.size() , n = obstacleGrid[0].size();
vector<vector<int>> dp(m+1,vector<int>(n+1,0));
dp[0][1] = 1;
for(int i = 1 ; i <= m ; ++i)
for(int j = 1 ; j <= n ; ++j)
if(!obstacleGrid[i-1][j-1])
dp[i][j] = dp[i-1][j]+dp[i][j-1];
return dp[m][n];
}
};
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