Algorithm之Unique Path

经典排列组合与动态规划题

一、原题：

    // https://leetcode.com/problems/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?
    
    
    +---+---+---+---+---+---+---+---+
    |ROB|   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |end|
    +---+---+---+---+---+---+---+---+
    
    
    */
    

解法一

    /*
    
    // https://discuss.leetcode.com/topic/5623
    // java-dp-solution-with-complexity-o-n-m
    
    The assumptions are:

        - When (n==0||m==0) the function always returns 1 since the robot
          can't go left or up.
    
        - For all other cells. The result = uniquePaths(m-1,n) + uniquePaths(m,n-1)
    
    Therefore I populated the edges with 1 first and use DP to get the full 2-D array.
    
    */
    
    /*
        
        +----+----+----+----+----+----+----+----+
        | 1  | 1  | 1  | 1  | 1  | 1  | 1  | 1  |
        +----+----+----+----+----+----+----+----+
        | 1  | 2  | 3  | 4  | 5  | 6  | 7  | 8  |
        +----+----+----+----+----+----+----+----+
        | 1  | 3  | 6  | 10 | 15 | 21 | 28 | 36 |
        +----+----+----+----+----+----+----+----+
        | 1  | 4  | 10 | 20 | 35 | 56 | 84 |120 |
        +----+----+----+----+----+----+----+----+
    
    */


    // Time complexity: O(n^2)

    public int uniquePaths(int m, int n) {
        int[][] map = new int[m][n];
        
        for(int i = 0; i < m; i++){
            map[i][0] = 1;
        }
        
        for(int j = 0; j < n; j++){
            map[0][j] = 1;
        }
        
        for(int i = 1; i < m; i++){
            for(int j = 1; j < n; j++){
                map[i][j] = map[i-1][j] + map[i][j-1];
            }
        }
        
        return map[m-1][n-1]; // the finish star.
    }

解法二

    // Time complexity: O(n^2)

    public int uniquePaths(int m, int n) {
        int[]dp = new int[n];
        dp[0] = 1;
        for(int i = 0; i < m; i++)
            for(int j = 1; j < n; j++)
                dp[j] = dp[j] + dp[j-1];
        return dp[n-1]; // the finish star.
    }

解法三

   // Time complexity: O(n)
   /*
    https://discuss.leetcode.com/topic/52660
    // java-0ms-solution-with-explanations
    
    
    When we solve this problem, we should keep in mind that this is a 
    permutation and combination problem of high school level. 
    Therefore, we need not to use DP solution or recursive solution.
    
    Given m and n, there will be m+n-2 steps. 
    Among these m+n-2 steps, n-1 steps are towards right and m-1 steps are towards down.
    
    the question is changed to: Select m-1 steps from m+n-2 steps.
    So, there will be (m-1)C(m+n-2) solutions, which is the same as (n-1)C(m+n-2). 
    
    All we need is to write a program quickly calculating (m-1)C(m+n-2) or (n-1)C(m+n-2).
*/    
    
    public int uniquePaths(int m, int n) {
         long result = 1;
         int steps = m + n - 2;

         for(int i=0; i < Math.min(m-1,n-1); i++)
             result = result * (steps - i) / (i + 1); 
             
         return (int)result;
    }


/*
    注意： 

    result = result * (steps - i) / (i + 1);

    不要写成

    result *= (steps - i) / (i + 1);
   
    的形式，否则后者在做除法时丢位导致结果错误。


*/

解法四

    // Time complexity: O(n)
    /*
    
    https://discuss.leetcode.com/topic/52660
    // java-0ms-solution-with-explanations
    
    https://discuss.leetcode.com/topic/31724
    // java-solution-0ms-4lines
    
    
    When we solve this problem, we should keep in mind that this is a 
    permutation and combination problem of high school level. 
    Therefore, we need not to use DP solution or recursive solution.
    
    Given m and n, there will be m+n-2 steps. 
    Among these m+n-2 steps, n-1 steps are towards right and m-1 steps are towards down.
    
    the question is converted to: Select m-1 steps from m+n-2 steps.
    So, there will be (m-1)C(m+n-2) solutions, which is the same as (n-1)C(m+n-2). 
    
    All we need is to write a program quickly calculating (m-1)C(m+n-2) or (n-1)C(m+n-2).
    

    
    */

  /*
    
    基础知识：（从 n 个数中 取出 m 个数）
    
    排列公式： result = n!/(n-m)!
    组合公式： result = n!/(n-m)!/m!
    
  ------------------------------------------------------  
    
    该题是一个组合题：
    
        7, 3
        
        steps = 8
        
        n = 3
        
        result = 8!/(3!*(8-3)!)
        
        8 7 6   5 4 3 2
        - - - * - - - -  
        3 2 1   5 4 3 2
        
 
     */
    public int uniquePaths(int m, int n) {
        int steps = m + n - 2;
        int min = Math.min(m - 1, n - 1);

        long sum = 1, sub = 1;        
        for(int i = 0; i < min; i++){
            sum *= (steps - i);
            sub *= (min - i);
        }

        return (int)(sum/sub);
   }

/*
   注意：这种解法乘完再除，相乘的结果可能会造成溢位，而出错。
         所以只适合数比较小的情况。

*/

二、增加些难度

/*
https://leetcode.com/problems/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.
*/


public int uniquePathsWithObstacles(int[][] obstacleGrid) {
    int width = obstacleGrid[0].length;
    int[] dp = new int[width];
    dp[0] = 1;
    for (int[] row : obstacleGrid) {
        for (int j = 0; j < width; j++) {
            if (row[j] == 1)
                dp[j] = 0;
            else if (j > 0)
                dp[j] += dp[j - 1];
        }
    }
    return dp[width - 1];
}

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