【LeetCode & 劍指offer刷題】動態規划與貪婪法題9:Unique Paths(系列)


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Unique Paths(系列)

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 a t 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 7 x 3 grid. How many possible unique paths are there?
Note:   m   and   n   will be at most 100.
Example 1:
Input: m = 3, n = 2
Output: 3
Explanation:
From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
1. Right -> Right -> Down
2. Right -> Down -> Right
3. Down -> Right -> Right
Example 2:
Input: m = 7, n = 3
Output: 28

C++
 
//問題:街區路徑問題(與LCS最長公共子序列問題類似) 求機器人從起點到終點的路徑數量
//方法一:設P[i][j] 為到坐標(i,j)的路徑總數,則由於機器人 只能向右或者向下 ,故有P[i][j] = P[i - 1][j] + P[i][j - 1]
//O(n^2) O(m*n)
class Solution
{
public :
    int uniquePaths ( int m , int n )
    {
        vector < vector < int >> path ( m , vector < int >( n , 1 )); //(1,0)和(0,1)處初始化為1
      
        for ( int i = 1 ; i < m ; i ++) //i從1開始,掃描行
            for ( int j = 1 ; j < n ; j ++) //掃描列
                path [i][j] = path[i-1][j] + path[i][j-1];
      
        return path [ m - 1 ][ n - 1 ];
    }
};
/*
* 方法一改進
  用一個列向量存當前需要的元素即可,將空間復雜度優化至O(min(m,n))
* 例子
    1 1 1 1  1  1
    1 2 3 4  5  6
    1 3 6 10 15 21
dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
dp[i][j],dp[i-1][j] ---> dp[j]
dp[i][j-1] ---> dp[j-1]
*/
class Solution
{
public :
    int uniquePaths ( int m , int n )
    {
        vector < int > dp ( n , 1 ); //進一步改進可以選擇min(m,n)
       
        for ( int i = 1 ; i < m ; ++ i ) //掃描行
        {
            for ( int j = 1 ; j < n ; ++ j ) //掃描列
            {
                dp [ j ] += dp [ j - 1 ];
            }
        }
        return dp [ n - 1 ];
    }
};
 
63 .   Unique Paths II
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.
Note:   m   and   n   will be at most 100.
Example 1:
Input:
[
[0,0,0],
[0,1,0],
[0,0,0]
]
Output: 2
Explanation:
There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:
1. Right -> Right -> Down -> Down
2. Down -> Down -> Right -> Right

 
/*
問題:有障礙物的柵格中,機器人從起始點到終點的路徑數量
方法:動態規划
對於有障礙物的柵格中,dp值置0
*/
class Solution
{
public :
    int uniquePathsWithObstacles ( vector < vector < int >>& obstacleGrid )
    {
        if ( obstacleGrid . empty () || obstacleGrid [ 0 ]. empty () || obstacleGrid [ 0 ][ 0 ] == 1 )
            return 0 ; //矩陣為空或者起始點為障礙點時,退出
       
        int m = obstacleGrid . size ();
        int n = obstacleGrid [ 0 ]. size ();
        vector < vector < int >> dp ( m , vector < int >( n )); //初始化為0
       
        for ( int i = 0 ; i < m ; i ++) //從0開始掃描,因為任何點都有可能為障礙點(如(0,1)和(1,0))
        {
            for ( int j = 0 ; j < n ; j ++)
            {
                if ( obstacleGrid [ i ][ j ] == 1 ) dp [ i ][ j ] = 0 ; //有障礙的地方dp值置0
                else if ( i == 0 && j == 0 ) dp [ i ][ j ] = 1 ;    //起始點到起始點路徑數為1(若無障礙)
                else if ( i == 0 && j > 0 ) dp [ i ][ j ] = dp [ i ][ j - 1 ]; //第一行的處理
                else if ( i > 0 && j == 0 ) dp [ i ][ j ] = dp [ i - 1 ][ j ]; //第一列的處理
                else dp [ i ][ j ] = dp [ i - 1 ][ j ] + dp [ i ][ j - 1 ]; //其他情況的處理
            }
        }
        return dp . back (). back (); //返回末尾點
    }
};
 

 


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