Unique Paths II - LeetCode

题目：

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.

分析：

这道题目和Minimum Path Sum，Unique Paths（点击打开链接）的思想一样，都是直接动态规划就可以。在grid[a][b]这一块时，我们只能从grid[a][b-1]和grid[a-1][b]到这一块，而从第一块能到这一块的路径总共有多少呢？等于到grid[a][b-1]和grid[a-1][b]的路径数相加。而当一直推到a = 1 和 b=1 时，我们到grid[a][b]的路径 = grid[a][b-1]和grid[a-1][b] ， 即grid[1][0]和grid[1][0]，而这两个都等于1，因为只能从grid[0][0]到这两个点。至于最上面和最左面的边缘点，只能从grid[0][0]移动到，所以为1.

代码：

class Solution:
    # @param obstacleGrid, a list of lists of integers
    # @return an integer
    def uniquePathsWithObstacles(self, obstacleGrid):
        if not obstacleGrid:
            return 0
        if obstacleGrid[0][0]== 0:
            obstacleGrid[0][0]=1
        else:
            return 0
        m,n =len(obstacleGrid),len(obstacleGrid[0])
        for j in xrange(0,n):
            for i in xrange(0,m):
                if obstacleGrid[i][j] == 1 and (i != 0 or j != 0):
                     obstacleGrid[i][j] = 0
                elif i == 0 and j == 0:
                    continue
                elif i == 0:
                    obstacleGrid[i][j] = obstacleGrid[i][j-1]
                elif j == 0:
                    obstacleGrid[i][j] = obstacleGrid[i-1][j]
                else:
                    obstacleGrid[i][j] = obstacleGrid[i-1][j]+obstacleGrid[i][j-1]
        return obstacleGrid[m-1][n-1]

Unique Paths II - LeetCode

原文：http://blog.csdn.net/u010006643/article/details/44836193