题意:给出两个排好序的数组,求出前k个和最小的数对。
思路:归并排序的归并思路,每次取最小的数对。
class Solution {
public:
vector<pair<int, int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
vector<vector<int>> mp;
for(int i = 0; i < nums1.size(); ++ i) {
vector<int> t;
for(int j = 0; j < nums2.size(); ++ j) {
t.push_back(nums1[i] + nums2[j]);
}
mp.push_back(t);
}
vector<int> pos(nums1.size(), 0);
vector<pair<int, int>> re;
pair<int, int> nextP;
int total = nums1.size() * nums2.size();
while(k -- && total --) {
int next = 0;
int findMin = INT_MAX;
for(int i = 0; i < pos.size(); ++ i) {
if(pos[i] >= nums2.size()) continue;
if(findMin > mp[i][pos[i]]) {
findMin = mp[i][pos[i]];
next = i;
}
}
//cout << next << " " << pos[next] << endl;
nextP.first = nums1[next];
nextP.second = nums2[pos[next]];
re.push_back(nextP);
pos[next] ++;
}
return re;
}
};