Leetcode: Ugly Number II
2015-08-20 23:18
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Write a program to find the
Ugly numbers are positive numbers whose prime factors only include
Note that
保存丑数序列,维护*2, *3 ,*5的三个队列,采用归并排序的思想选择最小的为下一个丑数。注意有可能某个丑数同时来自不同的队列。
class Solution {
public:
int nthUglyNumber(int n) {
vector<int> nums(n, 1);
int pos2 = 0, pos3 = 0, pos5 = 0;
for (int i = 1; i < n; ++i) {
nums[i] = min(min(nums[pos2] * 2, nums[pos3] * 3), nums[pos5] * 5);
if (nums[i] == nums[pos2] * 2) {
++pos2;
}
if (nums[i] == nums[pos3] * 3) {
++pos3;
}
if (nums[i] == nums[pos5] * 5) {
++pos5;
}
}
return nums[n-1];
}
};
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, 12is the sequence of the first
10ugly numbers.
Note that
1is typically treated as an ugly number.
保存丑数序列,维护*2, *3 ,*5的三个队列,采用归并排序的思想选择最小的为下一个丑数。注意有可能某个丑数同时来自不同的队列。
class Solution {
public:
int nthUglyNumber(int n) {
vector<int> nums(n, 1);
int pos2 = 0, pos3 = 0, pos5 = 0;
for (int i = 1; i < n; ++i) {
nums[i] = min(min(nums[pos2] * 2, nums[pos3] * 3), nums[pos5] * 5);
if (nums[i] == nums[pos2] * 2) {
++pos2;
}
if (nums[i] == nums[pos3] * 3) {
++pos3;
}
if (nums[i] == nums[pos5] * 5) {
++pos5;
}
}
return nums[n-1];
}
};
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