365天挑战LeetCode1000题——Day 108 优势洗牌

870. 优势洗牌

代码实现

class Solution {
public:
    vector<int> advantageCount(vector<int>& nums1, vector<int>& nums2) {
        int n = nums1.size();
        sort(nums1.begin(), nums1.end(), less<int>());
        priority_queue<pair<int, int>,
        vector<pair<int, int>>, greater<pair<int, int>>> pq;
        for (int i = 0; i < n; i++) {
            pq.push({nums2[i], i});
        }
        vector<int> ans(n, -1);
        queue<int> res;
        int j = 0;
        for (int i = 0; i < n && j < n; i++, j++) {
            auto [num, pos] = pq.top();
            pq.pop();
            while (j < n && num >= nums1[j]) {
                res.push(nums1[j]);
                j++;
            }
            if (j == n) {
                break;
            }
            ans[pos] = nums1[j];
        }
        while (j < n) {
            res.push(nums1[j]);
            j++;
        }
        for (int i = 0; i < n; i++) {
            if (ans[i] == -1) {
                ans[i] = res.front();
                res.pop();
            }
        }
        return ans;
    }
};