Advantage Shuffle LeetCode solution in C++

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class Solution {
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 public:
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  vector<int> advantageCount(vector<int> &nums1, vector<int> &nums2) {
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    int n1 = nums1.size();
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    int n2 = nums2.size();
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    multiset<int> s(nums1.begin(), nums1.end());
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    for (int i = 0; i < n2; i++) {
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      int val = nums2[i];
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      auto it = s.upper_bound(val);
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      if (it != s.end()) {
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        nums1[i] = *it;
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        s.erase(it);
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      } else {
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        nums1[i] = -1;
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      }
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    }
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    for (int i = 0; i < n1; i++) {
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      if (nums1[i] == -1) {
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        nums1[i] = *s.begin();
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        s.erase(s.begin());
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      }
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    }
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    return nums1;
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  }
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};