這個應該是微軟的一道面試題,首先需要明白醜數的概念。1是一個特殊的醜數,其次只含有2,3,5質因數的數被定義爲醜數。
關於解題的思路可以參考:
http://www.geeksforgeeks.org/ugly-numbers/
1到N的醜數爲 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, … ;
可以分成如下三組:
(1) 1×2, 2×2, 3×2, 4×2, 5×2, …
(2) 1×3, 2×3, 3×3, 4×3, 5×3, …
(3) 1×5, 2×5, 3×5, 4×5, 5×5, …
只需要保證每次迭代的過程中選取最小的醜數(或者說是質因數),可以理解爲後面的數字是前面的數組針對2,3,5的乘積。
C++可以AC的代碼如下:
//求三個數的最小值
int min(int a, int b, int c){
int minNum = a > b ? b : a;
return minNum > c ? c : minNum;
}
//求第N個醜數
int nthUglyNumber(int n) {
int *ugly = new int[n];
memset(ugly, 0, sizeof(int) * n);
ugly[0] = 1;
int factor2 = 2, factor3 = 3, factor5 = 5;
int index2, index3, index5;
index2 = index3 = index5 = 0;
for(int i=1; i<n; i++){
int minNum = min(factor2, factor3, factor5);
ugly[i] = minNum;
if(factor2 == minNum)
factor2 = 2 * ugly[++index2];
if(factor3 == minNum)
factor3 = 3 * ugly[++index3];
if(factor5 == minNum)
factor5 = 5 * ugly[++index5];
}
return ugly[n-1];
}