leetcode 之 Unique Paths

Unique Paths

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 at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).

How many possible unique paths are there?

Above is a 3 x 7 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

思路：典型的动态规划，dp[i][j]表示到matrix[i][j]的路径个数。则dp[i][j] = dp[i-1][j] + dp[i][j-1]。

int uniquePaths(int m, int n) {
	if(m <=0 || n <= 0)return 0;
	vector<vector<int> > dp(m);
	int i,j;
	for(i=0;i<m;i++)
	{
		vector<int> tmp(n,1);//至少一条
		dp[i] = tmp;
	}
	for(i = 1;i < m;i++)
	{
		for(j = 1;j < n;j++)
		{
			dp[i][j] = dp[i-1][j] + dp[i][j-1];
		}
	}
	return dp[m-1][n-1];
}

Unique Paths II

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.

思路：和上面类似，仅仅是当obstacleGrid[i][j] == 0时要把dp[i][j]=0。表示此路不通，初始化时也要注意

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
    	int rows = obstacleGrid.size();
    	if(rows <= 0)return 0;
    	int cols = obstacleGrid[0].size();
    	if(cols <= 0)return 0;
    	vector<vector<int> > dp(rows);
    	int i,j;
    	for(i = 0;i < rows;i++)
    	{
    		vector<int> tmp(cols);
    		dp[i] = tmp;
    	}
    	dp[0][0] = obstacleGrid[0][0] == 0 ? 1 : 0;
    	for(i = 1;i < rows;i++)dp[i][0] = obstacleGrid[i][0] == 0 ? dp[i-1][0] : 0;//当为0时，不能简单的初始化为1。要初始化为前面的值，由于可能被前面挡住了
    	for(j = 1;j < cols;j++)dp[0][j] = obstacleGrid[0][j] == 0 ?
 dp[0][j-1] : 0;
    	for(i = 1;i < rows;i++)
    	{
    		for(j = 1;j < cols;j++)
    		{
    			dp[i][j] = obstacleGrid[i][j] == 0 ? dp[i-1][j] + dp[i][j-1] : 0;
    		}
    	}
    	return dp[rows-1][cols-1];
    }
};