Ugly Number II

TLE解法：

public class Solution {public int nthUglyNumber(int n) {int count = 0;int i = 1;while(count<n){if(isugly(i)){count++;}i++;}return i;}public boolean isugly(int n){if(n == 1)return true;while(n%2==0)n /= 2;while(n%3 == 0) n/=3;while(n%5==0)n/=5;if(n == 1)return true;elsereturn false;}
}

AC解法：

可以将丑数分成三个队列

l1:1*2，2*2,3*2,4*2,5*2,6*2,8*2……

l2:1*3，2*3,3*3,4*3,5*3,6*3,8*3……

l3:1*5，2*5,3*5,4*5,5*5,6*5,8*5……

每次从三个队列中拿出队头的数比较，最小的数为当前ith的丑数，循环操作直到为i<n,就找到了nth的丑数。代码：

public class Solution {public int nthUglyNumber(int n) {int r = 0;List <Integer> l1 = new LinkedList<Integer>();List <Integer> l2 = new LinkedList<Integer>();List <Integer> l3 = new LinkedList<Integer>();l1.add(1);l2.add(1);l3.add(1);for(int i = 0;i<n;i++){r = l1.get(0)<l2.get(0)?(l1.get(0)<l3.get(0)?l1.get(0):l3.get(0)):(l2.get(0)<l3.get(0)?l2.get(0):l3.get(0));if(l1.get(0)==r)l1.remove(0);if(l2.get(0)==r)l2.remove(0);if(l3.get(0)==r)l3.remove(0);l1.add(r*2);l2.add(r*3);l3.add(r*5);}return r;}
}

转载于:https://www.cnblogs.com/92sunqing/p/5017988.html

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