leetcode 368. Largest Divisible Subset
2017-12-06 19:19
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368. Largest Divisible Subset
Given a set of distinct positive integers, find the largest subset such that every pair (Si, Sj)
of elements in this subset satisfies: Si % Sj =
0 or Sj % Si = 0.
If there are multiple solutions, return any subset is fine.
Example 1:
Example 2:
class Solution {
public:
vector<int> largestDivisibleSubset(vector<int>& nums)
{
if (nums.empty()) return vector<int> ();
sort(nums.begin(), nums.end());
vector<int> count(nums.size(), 1); //记录连续的个数
vector<int> last(nums.size(), -1); //记录上一个连续的位置
for (int i = 0; i < nums.size(); i++)
{
for (int j = i + 1; j < nums.size(); j++)
{
if ( nums[j] % nums[i] == 0 && count[j] < count[i] + 1)
{
count[j] = count[i] + 1;
last[j] = i;
}
}
}
int maxcount = 0;
int maxpos; //找到满足要求最大的那个数的位置
for (int i = 0; i < nums.size(); i++)
{
if (count[i] > maxcount)
{
maxcount = count[i];
maxpos = i;
}
}
vector<int> ret(1, nums[maxpos]);
while (last[maxpos] != -1) //通过last数组往前找满足要求的数
{
maxpos = last[maxpos];
ret.push_back(nums[maxpos]);
}
return ret;
}
};
Given a set of distinct positive integers, find the largest subset such that every pair (Si, Sj)
of elements in this subset satisfies: Si % Sj =
0 or Sj % Si = 0.
If there are multiple solutions, return any subset is fine.
Example 1:
nums: [1,2,3] Result: [1,2] (of course, [1,3] will also be ok)
Example 2:
nums: [1,2,4,8] Result: [1,2,4,8]
class Solution {
public:
vector<int> largestDivisibleSubset(vector<int>& nums)
{
if (nums.empty()) return vector<int> ();
sort(nums.begin(), nums.end());
vector<int> count(nums.size(), 1); //记录连续的个数
vector<int> last(nums.size(), -1); //记录上一个连续的位置
for (int i = 0; i < nums.size(); i++)
{
for (int j = i + 1; j < nums.size(); j++)
{
if ( nums[j] % nums[i] == 0 && count[j] < count[i] + 1)
{
count[j] = count[i] + 1;
last[j] = i;
}
}
}
int maxcount = 0;
int maxpos; //找到满足要求最大的那个数的位置
for (int i = 0; i < nums.size(); i++)
{
if (count[i] > maxcount)
{
maxcount = count[i];
maxpos = i;
}
}
vector<int> ret(1, nums[maxpos]);
while (last[maxpos] != -1) //通过last数组往前找满足要求的数
{
maxpos = last[maxpos];
ret.push_back(nums[maxpos]);
}
return ret;
}
};
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