leetcode egg drop java

topic:

You will get  K eggs, and can be used from one  1 to the  N  total  N story building.

Each egg features are the same, if an egg is broken, you can not then it fall.

You know there is floor  F to meet  0 <= F <= N any of the above  F eggs are broken down in the floor, from  F falling lower than its floor or floor eggs will not be broken.

Every move , you can take an egg (if you have a complete egg) and take it from any floor  X dropped (to meet  1 <= X <= N).

Your goal is to know exactly what  F the value is.

Regardless  F of how the initial value, you determine  F the minimum number of moves value is how much?

 

Example 1:

Input: K = 1, N = 2 
Output: 2 
Explanation: 
Egg Drop from one floor. If it is broken, we certainly know that F = 0. 
Otherwise, the eggs fall from the second floor. If it is broken, we certainly know that F = 1. 
If it is not broken, then we know for sure F = 2. 
Therefore, in the worst case we need to move F 2 times to determine how much.

Example 2:

Input: K = 2, N = 6 
Output: 3

Example 3:

Input: K = 3, N = 14 
Output: 4

prompt:

1. 1 <= K <= 100
2. 1 <= N <= 10000

Ideas:

class Solution {
     public  int superEggDrop ( int eggNum, int floorNum) {
         // eggNum represents a number of eggs, floorNum represents the number of floors 
        IF (eggNum == 0 )
             return 0 ;
         IF (eggNum ==. 1 )
             return floorNum;
         // DP [I ] [j] denotes the i-th egg up to the j measured in step floors 
        int [] [] DP = new new  int [floorNum +. 1] [eggNum +. 1 ]; 
        DP [ 0] [0] = 0 ;
         for ( int I =. 1; I <= floorNum; ++ I) {
            DP [I] [ 0] = 0 ;
             for ( int J =. 1; J <= eggNum; ++ J) {
                 // eggs broken eggs are not broken 
                dp [i] [j] = dp [i - 1] [j ] + DP [I -. 1] [J -. 1] +. 1 ;
                 IF (DP [I] [J]> = floorNum) 
                     return I; 
            } 
        } 
        return floorNum; 
    } 
}