LintCode -- 不同的路径 II

LintCode -- unique-paths-ii(不同的路径 II)

原题链接：http://www.lintcode.com/zh-cn/problem/unique-paths-ii/

跟进“不同的路径”：

现在考虑网格中有障碍物，那样将会有多少条不同的路径？

网格中的障碍和空位置分别用1和0来表示。

样例

如下所示在3x3的网格中有一个障碍物：
[
[0,0,0],
[0,1,0],
[0,0,0]
]
一共有2条不同的路径从左上角到右下角。

注意

m和n均不超过100

分析：

dp[ i ][ j ] = dp [ i+1 ][ j ] + dp[ i ][ j+1 ]

**** 时间复杂度 O（n*m）， 空间复杂度 O（n*m） ****

代码（C++、Java、Python）：

class Solution {
public:
/**
* @param obstacleGrid: A list of lists of integers
* @return: An integer
*/
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
// write your code here
int m = obstacleGrid.size();
int n = obstacleGrid[0].size();
vector<vector<int>> dp;
dp.resize(m+1);
for (int i = 0; i < m+1; i++){
dp[i].resize(n+1);
dp[i]
= 0;
}
for (int i = 0; i < n+1; i++)
dp[m][i] = 0;
for (int i = m-1; i >= 0; i--)
for (int j = n-1; j >= 0; j--){
if (obstacleGrid[i][j] == 1)
dp[i][j] = 0;
else if (i == m-1 && j == n-1) dp[i][j] = 1;
else dp[i][j] = dp[i+1][j] + dp[i][j+1];
}
return dp[0][0];
}
};

public class Solution {
/**
* @param obstacleGrid: A list of lists of integers
* @return: An integer
*/
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
// write your code here
int m = obstacleGrid.length;
int n = obstacleGrid[0].length;
int [][] dp = new int [m+1][n+1];
for (int i = m-1; i >= 0; i--)
for (int j = n-1; j >= 0; j--){
if (obstacleGrid[i][j] == 1) dp[i][j] = 0;
else if (i == m-1 && j == n-1) dp[i][j] = 1;
else dp[i][j] = dp[i+1][j] + dp[i][j+1];
}
return dp[0][0];
}
}

class Solution:
"""
@param obstacleGrid: An list of lists of integers
@return: An integer
"""
def uniquePathsWithObstacles(self, obstacleGrid):
# write your code here
m = len(obstacleGrid)
n = len(obstacleGrid[0])
dp = [[0 for i in range(n+1)] for j in range(m+1)]
for i in list(reversed(range(m))):
for j in list(reversed(range(n))):
if obstacleGrid[i][j] == 1:
dp[i][j] = 0
elif i == m-1 and j == n-1:
dp[i][j] = 1
else:
dp[i][j] = dp[i+1][j] + dp[i][j+1]
return dp[0][0]