Lintcode 4 Ugly Number II

原题如下：

Ugly number is a number that only have factors 2, 3 and 5.

Design an algorithm to find the nth ugly number. The first 10 ugly numbers are 1, 2, 3, 4, 5, 6, 8, 9, 10, 12...

Notice

Note that 1 is typically treated as an ugly number.

Example

If n=9, return 10.

题目的意思是看丑陋数，意思是只能被2,3,5相除的数，刚开始审题的时候没有看到题意中的only，一直卡在63%上，解题的方法很多，刚开始的做法是从一开始进行判断，直到n为止，但是这种方法十分耗时，因此会超时。

附上代码如下：

public class Solution {/*** 求第N个丑数* @param n* @return 第N个丑数*/public int nthUglyNumber(int n) {if (n <= 1) {return 1;}int counter = 0;for (int i = 1; ; i++) {if (isUgly(i)) {counter++;if (counter == n) {return i;}} }}/*** 判断数字是否为丑数* @param num 被判断数字* @return true：丑数，false：非丑数*/public boolean isUgly(int num) {if (num <= 0) {return false;}while (num % 2 == 0) num /= 2;while (num % 3 == 0) num /= 3;while (num % 5 == 0) num /= 5;if (num == 1) {return true;} else {return false;}}
}

后期加工的代码如下：

public class Solution {/*** 求第N个丑数* @param n* @return 第N个丑数*/public int nthUglyNumber(int n) {int[] uglyNumbers = new int[n];uglyNumbers[0] = 1;int idx2 = 0;int idx3 = 0;int idx5 = 0;int counter = 1;while (counter < n) {int min = minOf(uglyNumbers[idx2] * 2, uglyNumbers[idx3] * 3, uglyNumbers[idx5] * 5);if (min == uglyNumbers[idx2] * 2) {idx2++;}if (min == uglyNumbers[idx3] * 3) {idx3++;}if (min == uglyNumbers[idx5] * 5) {idx5++;}uglyNumbers[counter] = min;counter++;}return uglyNumbers[n - 1];}/*** 求三个数字中最小的数字* @param a 数字a* @param b 数字b* @param c 数字c* @return a、b、c中最小的数字*/private int minOf(int a, int b, int c) {int temp = a < b ? a : b;return temp < c ? temp : c; }
}

为了方便理解这段代码，在这段代码里加入了System.out.println函数用于将结果输出到控制台。解除这些注释行，并指定输入的n为15，执行函数时输出到控制台的结果如下：

uglyNumbers[0]:1
-----------
idx2:0;ugly[idx2]:1
idx3:0;ugly[idx3]:1
idx5:0;ugly[idx5]:1
idx2:0;idx3:0;idx5:0
min==ugly[idx2]*2:2
idx2:0→1
uglyNumbers[1]:2
-----------
idx2:1;ugly[idx2]:2
idx3:0;ugly[idx3]:1
idx5:0;ugly[idx5]:1
idx2:1;idx3:0;idx5:0
min==ugly[idx3]*3:3
idx3:0→1
uglyNumbers[2]:3
-----------
idx2:1;ugly[idx2]:2
idx3:1;ugly[idx3]:2
idx5:0;ugly[idx5]:1
idx2:1;idx3:1;idx5:0
min==ugly[idx2]*2:4
idx2:1→2
uglyNumbers[3]:4
-----------
idx2:2;ugly[idx2]:3
idx3:1;ugly[idx3]:2
idx5:0;ugly[idx5]:1
idx2:2;idx3:1;idx5:0
min==ugly[idx5]*5:5
idx5:0→1
uglyNumbers[4]:5
-----------
idx2:2;ugly[idx2]:3
idx3:1;ugly[idx3]:2
idx5:1;ugly[idx5]:2
idx2:2;idx3:1;idx5:1
min==ugly[idx2]*2:6
idx2:2→3
min==ugly[idx3]*3:6
idx3:1→2
uglyNumbers[5]:6
-----------
idx2:3;ugly[idx2]:4
idx3:2;ugly[idx3]:3
idx5:1;ugly[idx5]:2
idx2:3;idx3:2;idx5:1
min==ugly[idx2]*2:8
idx2:3→4
uglyNumbers[6]:8
-----------
idx2:4;ugly[idx2]:5
idx3:2;ugly[idx3]:3
idx5:1;ugly[idx5]:2
idx2:4;idx3:2;idx5:1
min==ugly[idx3]*3:9
idx3:2→3
uglyNumbers[7]:9
-----------
idx2:4;ugly[idx2]:5
idx3:3;ugly[idx3]:4
idx5:1;ugly[idx5]:2
idx2:4;idx3:3;idx5:1
min==ugly[idx2]*2:10
idx2:4→5
min==ugly[idx5]*5:10
idx5:1→2
uglyNumbers[8]:10
-----------
idx2:5;ugly[idx2]:6
idx3:3;ugly[idx3]:4
idx5:2;ugly[idx5]:3
idx2:5;idx3:3;idx5:2
min==ugly[idx2]*2:12
idx2:5→6
min==ugly[idx3]*3:12
idx3:3→4
uglyNumbers[9]:12
-----------
idx2:6;ugly[idx2]:8
idx3:4;ugly[idx3]:5
idx5:2;ugly[idx5]:3
idx2:6;idx3:4;idx5:2
min==ugly[idx3]*3:15
idx3:4→5
min==ugly[idx5]*5:15
idx5:2→3
uglyNumbers[10]:15
-----------
idx2:6;ugly[idx2]:8
idx3:5;ugly[idx3]:6
idx5:3;ugly[idx5]:4
idx2:6;idx3:5;idx5:3
min==ugly[idx2]*2:16
idx2:6→7
uglyNumbers[11]:16
-----------
idx2:7;ugly[idx2]:9
idx3:5;ugly[idx3]:6
idx5:3;ugly[idx5]:4
idx2:7;idx3:5;idx5:3
min==ugly[idx2]*2:18
idx2:7→8
min==ugly[idx3]*3:18
idx3:5→6
uglyNumbers[12]:18
-----------
idx2:8;ugly[idx2]:10
idx3:6;ugly[idx3]:8
idx5:3;ugly[idx5]:4
idx2:8;idx3:6;idx5:3
min==ugly[idx2]*2:20
idx2:8→9
min==ugly[idx5]*5:20
idx5:3→4
uglyNumbers[13]:20
-----------
idx2:9;ugly[idx2]:12
idx3:6;ugly[idx3]:8
idx5:4;ugly[idx5]:5
idx2:9;idx3:6;idx5:4
min==ugly[idx2]*2:24
idx2:9→10
min==ugly[idx3]*3:24
idx3:6→7
uglyNumbers[14]:24
-----------
return:24

摘抄网上大佬的其他方法：

HashMap + PriorityQueue(MinHeap)：将2，3，5入队，每次取出堆顶元素（最小值），依次和3个factor相乘，将所得结果入队。用HashMap记录入队元素，入过队的元素不能再次入队。

代码如下：

HashMap + Heap O(nlogn)：

class Solution {/*** @param n an integer* @return the nth prime number as description.*/public int nthUglyNumber(int n) {// Write your code here//HashMap + Heap O(nlogn)if(n <= 0){return 0;}PriorityQueue<Long> Q = new PriorityQueue<Long>();HashMap<Long, Boolean> inQ = new HashMap<Long, Boolean>();Long[] prime = new Long[3];prime[0] = Long.valueOf(2);prime[1] = Long.valueOf(3);prime[2] = Long.valueOf(5);for(int i = 0; i < 3; i++){Q.add(prime[i]);inQ.put(prime[i], true);}//每次取最小堆（优先队列）顶的元素，取n-1次，同时将取出的数和3个factor相乘的结果加入队列（如果本来已经在队列中则不用加）Long number = Long.valueOf(1);for(int i = 1; i < n; i++){number = Q.poll();for(int j = 0; j < 3; j++){if(!inQ.containsKey(number * prime[j])){Q.add(number * prime[j]);inQ.put(number * prime[j], true);}}}//Long转换为Intreturn number.intValue();}
}

另有简便方法：

class Solution {
public:
int nthUglyNumber(int n) {
vector<int> res(1, 1);
int i2 = 0, i3 = 0, i5 = 0;
while (res.size() < n) {
int m2 = res[i2] * 2, m3 = res[i3] * 3, m5 = res[i5] * 5;
int mn = min(m2, min(m3, m5));
if (mn == m2) ++i2;
if (mn == m3) ++i3;
if (mn == m5) ++i5;
res.push_back(mn);
}
return res.back();
}
};

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