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_264_nthUglyNumber.java
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_264_nthUglyNumber.java
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package pp.arithmetic.leetcode;
import java.util.HashMap;
/**
* Created by wangpeng on 2019-05-09.
* 264. 丑数 II
* <p>
* 编写一个程序,找出第 n 个丑数。
* <p>
* 丑数就是只包含质因数 2, 3, 5 的正整数。
* <p>
* 示例:
* <p>
* 输入: n = 10
* 输出: 12
* 解释: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 是前 10 个丑数。
* 说明:
* <p>
* 1 是丑数。
* n 不超过1690。
*
* @see <a href="https://leetcode-cn.com/problems/ugly-number-ii/">ugly-number-ii</a>
*/
public class _264_nthUglyNumber {
public static void main(String[] args) {
_264_nthUglyNumber nthUglyNumber = new _264_nthUglyNumber();
System.out.println(nthUglyNumber.nthUglyNumber(10));
long start = System.currentTimeMillis();
System.out.println(nthUglyNumber.nthUglyNumber(431));
long end = System.currentTimeMillis();
System.out.println("循环法计算431耗时:" + (end - start));
start = System.currentTimeMillis();
System.out.println(nthUglyNumber.nthUglyNumber2(431));
end = System.currentTimeMillis();
System.out.println("三指针法计算431耗时:" + (end - start));
}
/**
* 解题二:动态规划+三指针
* dp保存按序排列的丑数,三指针分别是*2,*3,*5,找出下一个丑数
*
* @param n
* @return
*/
public int nthUglyNumber2(int n) {
int[] dp = new int[n];
dp[0] = 1;
int i2 = 0, i3 = 0, i5 = 0;
for (int i = 1; i < n; i++) {
int min = Math.min(dp[i2] * 2, Math.min(dp[i3] * 3, dp[i5] * 5));
if (min == dp[i2] * 2) i2++;
if (min == dp[i3] * 3) i3++;
if (min == dp[i5] * 5) i5++;
dp[i] = min;
}
return dp[n - 1];
}
private HashMap<Integer, Boolean> map = new HashMap<>();
/**
* 丑数求解过程:首先除2,直到不能整除为止,然后除5到不能整除为止,然后除3直到不能整除为止。
* 最终判断剩余的数字是否为1,如果是1则为丑数,否则不是丑数
* <p>
* 解题思路:
* 从1开始遍历,按丑数求解过程找出满足条件的第n个丑数(提交超时)
* 思路优化(如何利用之前的计算)
*
* @param n
* @return
*/
public int nthUglyNumber(int n) {
map.put(1, true);
int uglyCount = 1;
int retVal = 1;
for (int i = 2; i < Integer.MAX_VALUE; i++) {
if (uglyCount >= n) {
break;
}
boolean isUgly = isUglyNumber(i);
if (isUgly) {
map.put(i, true);
uglyCount++;
retVal = i;
}
}
return retVal;
}
private boolean isUglyNumber(int num) {
while (num % 2 == 0) {
num = num / 2;
if (map.containsKey(num)) return true;
}
while (num % 5 == 0) {
num = num / 5;
if (map.containsKey(num)) return true;
}
while (num % 3 == 0) {
num = num / 3;
if (map.containsKey(num)) return true;
}
return num == 1;
}
}