Leetcode之Ugly Number II

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本文介绍了一种高效算法，用于查找序列中的第N个丑数。丑数定义为仅包含2、3、5作为质因数的正整数。通过使用动态规划方法，我们能够按顺序生成丑数序列，并找到指定位置的丑数。

题目：

Write a program to find the n-th ugly number.

Ugly numbers are positive numbers whose prime factors only include 2, 3, 5.

Example:

Input: n = 10
Output: 12
Explanation: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 is the sequence of the first 10 ugly numbers.

Note:

1 is typically treated as an ugly number.
n does not exceed 1690.

代码：

class Solution {
public:
    int nthUglyNumber(int n) {
        if(n <= 0) return false; // get rid of corner cases 
        if(n == 1) return true; // base case
        int t2 = 0, t3 = 0, t5 = 0; //pointers for 2, 3, 5
        vector<int> k(n);
        k[0] = 1;
        for(int i  = 1; i < n ; i ++)
        {
            k[i] = min(k[t2]*2,min(k[t3]*3,k[t5]*5));
            if(k[i] == k[t2]*2) t2++; 
            if(k[i] == k[t3]*3) t3++;
            if(k[i] == k[t5]*5) t5++;
        }
        return k[n-1];
    }
};