LeetCode -- Ugly Number II

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public int NthUglyNumber(int n) 
{
	var hash = new Dictionary<int, int>();
	hash.Add(1 , 1);
	hash.Add(2 , 2);
	hash.Add(3 , 3);
	hash.Add(4 , 4);
	hash.Add(5 , 5);
	if(hash.ContainsKey(n)){
		return hash[n];
	}
	var count = 5;
	var n1 = 6;
	while(count < n){
		if(n1 % 2 == 0 && hash.ContainsKey(n1/2) ||
		   n1 % 3 == 0 && hash.ContainsKey(n1/3) ||
		   n1 % 5 == 0 && hash.ContainsKey(n1/5)){
			hash.Add(n1, count);
			count ++;
		}
		n1++;
	}
	
	return hash.Keys.Last();
}

public class Solution {
    public int NthUglyNumber(int n) 
    {
    	var i2 = 1;
    	var i3 = 1;
    	var i5 = 1;
    	var uglies = new List<int>();
    	uglies.Add(1);
    	while (uglies.Count < n) {
    		var v2 = uglies[i2-1] * 2;
    		var v3 = uglies[i3-1] * 3;
    		var v5 = uglies[i5-1] * 5;
    		
    		int min = Math.Min(v2, Math.Min(v3, v5));
    		if (min == v2) {
    			i2++;
    		}
    		if (min == v3){
    			i3++;
    		}
    		if (min == v5) {
    			i5++;
    		}
    		uglies.Add(min);
    	}
    	return uglies.Last();
    }
}

LeetCode -- Ugly Number II

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原文地址：http://blog.csdn.net/lan_liang/article/details/49885465