【刷题-LeetCode】264. Ugly Number II
- Ugly Number II
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.
解法1
暴力搜索。假设已经知道了前n个丑数\(a_1, a_2, ..., a_n\),求第n+1个丑数,则有:
s.t\quad a_{n+1} > a_n
\]
class Solution {
public:
int nthUglyNumber(int n) {
vector<int>u_n{1};
int prime[3] = {2, 3, 5};
while(u_n.size() < n){
int flag = 0;
int cur_n = INT_MAX;
for(int i = u_n.size() - 1; i >= 0; --i){
for(int j = 0; j < 3; ++j){
if(u_n[i]*prime[j] > u_n.back()){
cur_n = min(cur_n, u_n[i]*prime[j]);
}else{
flag++;
}
}
if(flag == 3)break;
}
u_n.push_back(cur_n);
}
return u_n.back();
}
};
但是提交会超时。。。
解法2 对解法1进行改进。显然丑数数组是有序的,可以用二分查找完成,查找的问题描述为:
寻找第一次出现的满足 \(k\times a_i > a_n\)的\(a_i\)
搜索过程为:对于区间\([l, r]\)
- \(k\times a_{mid} > a_n\),则满足条件的肯定在\([l, mid]\)中
- \(k\times a_{mid} \leq a_n\),则满足条件的肯定在\([mid+1, r]\)中
typedef long long int LL;
class Solution {
public:
int nthUglyNumber(int n) {
vector<LL>u_n{1};
int prime[3] = {2, 3, 5};
while(u_n.size() < n){
LL cur_n = LLONG_MAX;
for(int j = 0; j < 3; ++j){
int idx = bin_search(u_n, prime[j]);
cur_n = min(cur_n, prime[j]*u_n[idx]);
}
u_n.push_back(cur_n);
}
return u_n.back();
}
int bin_search(vector<LL>&nums, int k){
int last_num = nums.back();
int l = 0, r = nums.size()-1;
while(l < r){
int mid = (l + r) / 2;
if(nums[mid]*k > last_num)r=mid;
else l = mid + 1;
}
return l;
}
};
解法3 注意到事实:如果\(a_n\)是丑数,则\(2a_n, 3a_n, 5a_n\)也是丑数
typedef long long int LL;
class Solution {
public:
int nthUglyNumber(int n) {
priority_queue<LL, vector<LL>, greater<LL>>q;
set<LL>s;
int prime[3] = {2, 3, 5};
q.push(1);
s.insert(1);
LL ans = q.top();
for(int i = 0; i < n; ++i){
ans = q.top();
q.pop();
s.erase(ans);
for(int j = 0; j < 3; ++j){
if(s.find(ans*prime[j]) == s.end()){
q.push(ans*prime[j]);
s.insert(ans*prime[j]);
}
}
}
return ans;
}
};
解法4 根据解法三种事实,利用动态规划
class Solution {
public:
int nthUglyNumber(int n) {
int pre2 = 0, pre3 = 0, pre5 = 0;
int nums[1690];
nums[0] = 1;
for(int i = 1; i < n; ++i){
int ugly = min(nums[pre2]*2, min(nums[pre3]*3, nums[pre5]*5));
nums[i] = ugly;
if(ugly % 2 == 0)pre2++;
if(ugly % 3 == 0)pre3++;
if(ugly % 5 == 0)pre5++;
}
return nums[n-1];
}
};
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