leetcode——Unique Paths & Unique PathsⅡ

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.

解析：可利用動态規劃解題，假設目的地為（i，j），則可以通過（i-1，j）或者（i，j-1）到達，是以可得：dp[i][j] = dp[i-1][j] + dp[i][j-1]。時間複雜度為O(m*n)，空間複雜度為O(m*n)， 代碼如下：

class Solution {
public:
	int uniquePaths(int m, int n) {
		int dp[m][n];
		for(int i=0; i<m; i++)	dp[i][0] = 1;
		for(int i=0; i<n; i++)	dp[0][i] = 1;
		for(int i=1; i<m; i++)
			for(int 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 

 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.

解析：加入障礙之後，顯然若格子（i，j）為障礙，則dp[i][j] = 0，此外初始化時進行處理即可，其餘與Unique Paths類似。時間複雜度為O(m*n)，空間複雜度為O(m*n)，代碼如下：

class Solution {
public:
	int uniquePathsWithObstacles(vector<vector<int> >& obstacleGrid) {
		int m = obstacleGrid.size(), n = obstacleGrid[0].size(), dp[m][n];
		int i = 0, j = 0;
		for(i=0; i<m && obstacleGrid[i][0]!=1; i++)	dp[i][0] = 1;
		for(; i<m; i++)	dp[i][0] = 0;
		for(j=0; j<n && obstacleGrid[0][j]!=1; j++)	dp[0][j] = 1;
		for(; j<n; j++)	dp[0][j] = 0;
		
		for(i=1; i<m; i++)
			for(j=1; j<n; j++)
			{
				if(obstacleGrid[i][j] == 1)	dp[i][j] = 0;
				else	dp[i][j] = dp[i-1][j] + dp[i][j-1];
			}
		return dp[m-1][n-1];
	}
};