70 lines
2.1 KiB
C++
70 lines
2.1 KiB
C++
// Source : https://leetcode.com/problems/super-ugly-number/
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// Author : Hao Chen
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// Date : 2017-01-02
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/***************************************************************************************
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*
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* Write a program to find the nth super ugly number.
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*
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* Super ugly numbers are positive numbers whose all prime factors are in the given
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* prime list
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* primes of size k. For example, [1, 2, 4, 7, 8, 13, 14, 16, 19, 26, 28, 32]
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* is the sequence of the first 12 super ugly numbers given primes
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* = [2, 7, 13, 19] of size 4.
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*
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* Note:
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* (1) 1 is a super ugly number for any given primes.
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* (2) The given numbers in primes are in ascending order.
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* (3) 0 k ≤ 100, 0 n ≤ 106, 0 primes[i]
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*
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* Credits:Special thanks to @dietpepsi for adding this problem and creating all test
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* cases.
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***************************************************************************************/
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// As the solution we have for the ugly number II problem
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//
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// int nthUglyNumber(int n) {
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//
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// int i=0, j=0, k=0;
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// vector<int> ugly(1,1);
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//
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// while(ugly.size() < n){
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// int next = min(ugly[i]*2, ugly[j]*3, ugly[k]*5);
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// if (next == ugly[i]*2) i++;
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// if (next == ugly[j]*3) j++;
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// if (next == ugly[k]*5) k++;
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// ugly.push_back(next);
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// }
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// return ugly.back();
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// }
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//
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// The logic of solution is exacly same for both., except that instead of 3 numbers you have k numbers to consider.
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//
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//
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//
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class Solution {
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public:
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int nthSuperUglyNumber(int n, vector<int>& primes) {
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vector<int> ugly(1, 1);
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int len = primes.size();
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vector<int> pos(len, 0);
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while( ugly.size() < n ) {
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int next = INT_MAX;
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for(int i=0; i<len; i++) {
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next = min(next, ugly[pos[i]] * primes[i]);
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}
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for(int i=0; i<len; i++) {
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if (next == ugly[pos[i]] * primes[i]) {
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pos[i]++;
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}
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}
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ugly.push_back(next);
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}
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return ugly.back();
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}
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};
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