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 程式師世界 >> 編程語言 >> C語言 >> C++ >> 關於C++ >> [LeetCode] Ugly Number II

[LeetCode] Ugly Number II

編輯:關於C++

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

Ugly numbers are positive numbers whose prime factors only include 2, 3, 5. For example, 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 is the sequence of the first 10 ugly numbers.

Note that 1 is typically treated as an ugly number.

Hint:

The naive approach is to call isUgly for every number until you reach the nth one. Most numbers are not ugly. Try to focus your effort on generating only the ugly ones. An ugly number must be multiplied by either 2, 3, or 5 from a smaller ugly number. The key is how to maintain the order of the ugly numbers. Try a similar approach of merging from three sorted lists: L1, L2, and L3. Assume you have Uk, the kth ugly number. Then Uk+1 must be Min(L1 * 2, L2 * 3, L3 * 5).

解題思路

見題目中的提示,主要就是維持一個丑數序列,一個丑數一定是序列中比它小的丑數乘以2,3或5的結果。

實現代碼

Java:

// Runtime: 9 ms
public class Solution {
    public int nthUglyNumber(int n) {
        int[] num = new int[n];
        num[0] = 1;
        int n2 = 0;
        int n3 = 0;
        int n5 = 0;
        int i = 1;
        while (i < n) {
            int ugly = Math.min(Math.min(num[n2] * 2, num[n3] * 3), num[n5] * 5);
            num[i++] = ugly;
            if (ugly == num[n2] * 2) {
                n2++;
            }
            if (ugly == num[n3] * 3) {
                n3++;
            }
            if (ugly == num[n5] * 5) {
                n5++;
            }
        }

        return num[n - 1];
    }
}

 

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