Unique Paths I 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 atany point in time. The robot is trying to reach the bottom-right corner of thegrid (marked 'Finish' in the diagram below).

        How many possible unique paths are there?

        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.

 

        思路：首先考虑没有障碍的情况，这是一个典型的斐波那契堆类型的问题：每一步都有两种选择，向右走或者向下走，数学表达式就是f(m, n) = f(m-1, n)+f(m, n-1)。所以可以很快写出一个递归版本的算法。

        但是对于较大的m和n，递归版本的空间复杂度和时间复杂度都很大。所以考虑非递归的实现。

        非递归的实现方法，从最后一行的最后一格开始，从右向左，从下向上依次求值，求值表达式为board[i][j] = board[i][j+1] + board[i+1][j]。最后一行和最后一列要特殊考虑。

        考虑添加了障碍物的情况，如果board[i][j]的初始值为1，表明此路不通，所以可以将其置为0即可。代码如下：

int uniquePathsWithObstacles(int**obstacleGrid, int obstacleGridRowSize, int obstacleGridColSize){     int i, j;    int res = 0;     for(i = obstacleGridRowSize-1;i >= 0; i--)    {        for(j = obstacleGridColSize-1;j >= 0; j--)        {            if(obstacleGrid[i][j]== 1)            {                obstacleGrid[i][j]= 0;                continue;            }            if(i == obstacleGridRowSize-1)            {                if(j == obstacleGridColSize-1)  obstacleGrid[i][j] =1;                else obstacleGrid[i][j]= obstacleGrid[i][j+1];            }            else if(j == obstacleGridColSize-1)            {                obstacleGrid[i][j]= obstacleGrid[i+1][j];            }               else            {                obstacleGrid[i][j]= obstacleGrid[i+1][j] + obstacleGrid[i][j+1];            }        }    }    res = obstacleGrid[0][0];       return res;}