田忌赛马背后的算法决策
870. 优势洗牌
class Solution {
public:
vector<int> advantageCount(vector<int>& nums1, vector<int>& nums2) {
int n = nums1.size();
priority_queue<pair<int, int>, vector<pair<int, int>>, function<bool(pair<int, int>, pair<int, int>)>> maxpq([](auto a, auto b){return a.second < b.second;});
for (int i = 0; i < n; i++) {
maxpq.emplace(i, nums2[i]);
}
sort(nums1.begin(), nums1.end());
vector<int> res(n);
int left = 0, right = n-1;
while (!maxpq.empty()) {
auto [i, maxval] = maxpq.top();
maxpq.pop();
if (maxval < nums1[right]) {
res[i] = nums1[right];
right--;
}
else {
res[i] = nums1[left];
left++;
}
}
return res;
}
};