leecode 解题总结：63. Unique Paths II

#include <iostream>
#include <stdio.h>
#include <vector>
using namespace std;
/*
问题:
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.

Note: m and n will be at most 100.

分析:同样是程序员面试金典的题目。设置了障碍。可以采用递归来做
可在计算的时候如果元素为1（障碍物），就设置该元素对应的路径为0来做

输入:
3(行数) 3(列数)
0 0 0
0 1 0
0 0 0

1 3
0 1 0
输出:
2
0

关键:
1 可在计算的时候如果元素为1（障碍物），就设置该元素对应的路径为0来做
初始化第一列,找到从上到下第一个障碍物，将该元素，以及该元素下面元素路径数都要设置为0
同理初始化第一行
*/

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
		if(obstacleGrid.empty())
		{
			return 0;
		}
		int row = obstacleGrid.size();//行数
		int col = obstacleGrid.at(0).size();//列数
		vector< vector<int> > paths(row  , vector<int> (col , 1));
		//初始化第一行和第一列元素，如果对应元素不是障碍物，对应路径数为1，否则为0
		//初始化第一列,找到从上到下第一个障碍物，将该元素，以及该元素下面元素路径数都要设置为0
		int i = 0;
		while(i < row)
		{
			//是障碍物，将该元素以及下面元素路径数都要设置为1
			if( 1 == obstacleGrid.at(i).at(0) )
			{
				while(i < row)
				{
					paths.at(i++).at(0) = 0;
				}
				break;
			}
			else
			{
				i++;
			}
		}
		//初始化第一行
		i = 0;
		while(i < col)
		{
			if( 1 == obstacleGrid.at(0).at(i) )
			{
				while(i < col)
				{
					paths.at(0).at(i++) = 0;
				}
			}
			else
			{
				i++;
			}
		}
		//接下来递推
		for(i = 1 ; i < row ; i++)
		{
			for(int j = 1 ; j < col ; j++)
			{
				if( obstacleGrid.at(i).at(j) != 1)
				{
					paths.at(i).at(j) = paths.at(i).at(j-1) + paths.at(i-1).at(j);
				}
				else
				{
					paths.at(i).at(j) = 0;
				}
			}
		}
		return paths.at(row - 1).at(col - 1);
    }
};


void process()
{
	 vector< vector<int> > nums;
	 int value;
	 int row;
	 int col;
	 Solution solution;
	 while(cin >> row >> col)
	 {
		 nums.clear();
		 for(int i = 0 ; i < row ; i++)
		 {
			 vector<int> num;
			 for(int j = 0 ; j < col; j++)
			 {
				 cin >> value;
				 num.push_back(value); 
			 }
			 nums.push_back(num);
		 }
		 int result = solution.uniquePathsWithObstacles(nums);
		 cout << result << endl;
	 }
}

int main(int argc , char* argv[])
{
	process();
	getchar();
	return 0;
}