leetcode_373 Find K Pairs with Smallest Sums

static bool cmp(pair<int, int> a, pair<int, int> b)
{
if (a.first + a.second < b.first + b.second)
return true;
return false;
}
// 方法1 暴力求解
vector< pair<int, int> > kSmallestPairs1(vector<int>& nums1, vector<int>& nums2, int k)
{
vector< pair<int, int> > result;
int size1 = (int)nums1.size();
int size2 = (int)nums2.size();
if (size1 <= 0 || size2 <= 0)
return result;
for (int i = 0; i < min(size1, k); i++)
{
for (int j = 0; j < min(size2, k); j++)
{
result.push_back(make_pair(nums1[i], nums2[j]));
}
}
// 对生成的结果进行排序
sort(result.begin(), result.end(), cmp);
// 删除重复的元素
if (result.size() > k)
result.erase(result.begin() + k, result.end());
return result;
}
// 方法2 借助multimap实现
vector< pair<int, int> > kSmallestPairs2(vector<int>& nums1, vector<int>& nums2, int k)
{
vector< pair<int, int> > result;
int size1 = nums1.size();
int size2 = nums2.size();
if (size1 <= 0 || size2 <= 0 || k <= 0)
return result;
multimap< int, pair<int, int> > mul;
for (int i = 0; i < min(size1, k); i++)
{
for (int j = 0; j < min(size2, k); j++)
{
int temp = nums1[i] + nums2[j];
mul.insert(make_pair(temp, make_pair(nums1[i], nums2[j])));
}
}
// 取出前k个元素
multimap< int, pair<int, int> >::iterator it;
for (it = mul.begin(); it != mul.end(); it++)
{
result.push_back(it->second);
k--;
if (k <= 0)
return result;
}
return result;
}
// 方法3：借助priority_queue实现
struct cmp1
{
bool operator()(pair<int, int> a, pair<int, int> b)
{
return a.first + a.second < b.first + b.second;
}
};
vector< pair<int, int> > kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k)
{
vector< pair<int, int> > result;
int size1 = nums1.size();
int size2 = nums2.size();
if (size1 <= 0 || size2 <= 0 || k <= 0)
return result;
priority_queue< pair<int, int>, vector< pair<int, int> >, cmp1> que;
for (int i = 0; i < min(size1, k); i++)
{
for (int j = 0; j < min(size2, k); j++)
{
if (que.size() < k)
{
que.push(make_pair(nums1[i], nums2[j]));
}
else
{
if (nums1[i] + nums2[j] < que.top().first + que.top().second)
{
que.push(make_pair(nums1[i], nums2[j]));
que.pop();
}
}
}

}
while (!que.empty())
{
result.push_back(que.top());
que.pop();
}
return result;
}