lintcode584- Drop Eggs II- medium

There is a building of n floors. If an egg drops from the k th floor or above, it will break. If it's dropped from any floor below, it will not break.

You're given m eggs, Find k while minimize the number of drops for the worst case. Return the number of drops in the worst case.

Example

Given m = 2, n = 100 return 14
Given m = 2, n = 36 return 8

 

动态规划+递归。用二元数组存储某鸡蛋某层所需的次数。迭代试扔第一个鸡蛋，在某层扔。1.扔碎了即转为鸡蛋少一个，楼层少一层的子问题。2.没扔碎即转化为鸡蛋没有少楼层少为上半层那么多的子问题。

 

思维方式见：http://datagenetics.com/blog/july22012/index.html的“Look see I can do three” 

 

 

public class Solution {
    /*
     * @param m: the number of eggs
     * @param n: the number of floors
     * @return: the number of drops in the worst case
     */
     
    public int dropEggs2(int m, int n) {
        // write your code here
        int[][] dp = new int[m + 1][n + 1];
        
        for (int i = 1; i <= m; i++){
            dp[i][0] = 0;
            dp[i][1] = 1;
        }
        
        for (int j = 1; j <= n; j++){
            dp[1][j] = j;
        }
        
        for (int i = 2; i <= m; i++){
            for (int j = 2; j <= n; j++){
                dp[i][j] = Integer.MAX_VALUE;
                for(int k = 1; k < j; k++){
                    dp[i][j] = Math.min(dp[i][j],
                                        1 + Math.max(dp[i - 1][k - 1], dp[i][j - k]));
                }
            }
        }
        return dp[m][n];
    }
}