Unique PathII.txt

problem：
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.

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solution：
思路一、用二维数组做记录
//（1）对于第1行，如果有障碍，则障碍处开始往后，方法数均为0，否则为1；
//（2）对于第1列，如果有障碍，则障碍处开始往下，方法数均为0，否则为1；
//（3）到达A[i][j]，如果A[i][j]有障碍，则方法数为0，否则，可以从A[i-1][j]到达A[i][j]，也可以从A[i][j-1]到达A[i][j]，可用的方法有A[i-1][j]+A[i][j-1]种。
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid)
{
  int m = obstacleGrid.size();
	if(m == 0)
		return 0;
	int n = obstacleGrid[0].size();
	if(n == 0)
		return 0;
	int dp[m][n];
	bool isBlocked = false;
	for(int c = 0; c < n; ++c)
	{
		if(isBlocked == false && obstacleGrid[0][c] == 1)
			isBlocked = true;
		if(isBlocked == false)
			dp[0][c] = 1;
		else
			dp[0][c] = 0;
	}
	isBlocked = false;
	for(int r = 0; r < m; ++r)
	{
		if(isBlocked == false && obstacleGrid[r][0] == 1)
			isBlocked = true;
		if(isBlocked == false)
			dp[r][0] = 1;
		else
			dp[r][0] = 0;
	}
	for (int r = 1; r < m; ++r)
		for (int c = 1; c < n; ++c)
		{
			if(obstacleGrid[r][c] == 0)
				dp[r][c] = dp[r - 1][c] + dp[r][c - 1];
			else 
				dp[r][c] = 0;
		}
	return dp[m - 1][n - 1];
}



思路二、用一维数组按列记录，从底往上，从右往左。
初始时最底行从右往左，每列初始为1直到遇到有障碍，则全为0；
接着从倒数第二行一直到第一行，如果没有障碍，则dp[i]=dp[i]+dp[i+1]，否则为0
////////////////////////////////////////////
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid)
{
	int m = obstacleGrid.size();
	int n = obstacleGrid[0].size();
	if(m == 0 || n == 0)
		return 0;
	vector<int> dp(n + 1, 0);
	for(int i = n - 1; i >= 0; --i)
		if(obstacleGrid[m - 1][i] == 0)
			dp[i] = 1;
	for(int j = m - 2; j >= 0; --j)
		for(int i = n - 1; i >= 0; --i)
			dp[i] = (obstacleGrid[j][i] == 1) ? 0 : dp[i]+dp[i+1];

	return dp[0];
}