题目：

题解：

class Solution {
public:
    vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
        int m = nums1.size();
        int n = nums2.size();
        auto count = [&](int target){
            long long cnt = 0;
            int start = 0;
            int end = n - 1;
            while (start < m && end >= 0) {
                if (nums1[start] + nums2[end] > target) {
                    end--;
                } else {
                    cnt += end + 1;
                    start++;
                }
            }
            return cnt;
        };

        /*二分查找第 k 小的数对和的大小*/
        int left = nums1[0] + nums2[0];
        int right = nums1.back() + nums2.back();
        int pairSum = right;
        while (left <= right) {
            int mid = left + ((right - left) >> 1);
            if (count(mid) < k) {
                left = mid + 1;
            } else {
                pairSum = mid;
                right = mid - 1;
            }
        }

        vector<vector<int>> ans;
        int pos = n - 1;
        /*找到小于目标值 pairSum 的数对*/
        for (int i = 0; i < m; i++) {
            while (pos >= 0 && nums1[i] + nums2[pos] >= pairSum) {
                pos--;
            }
            for (int j = 0; j <= pos && k > 0; j++, k--) {
                ans.push_back({nums1[i], nums2[j]});
            }
        }
        /*找到等于目标值 pairSum 的数对*/
        pos = n - 1;
        for (int i = 0; i < m && k > 0; i++) {
            int start1 = i;
            while (i < m - 1 && nums1[i] == nums1[i + 1]) {
                i++;
            }
            while (pos >= 0 && nums1[i] + nums2[pos] > pairSum) {
                pos--;
            }
            int start2 = pos;
            while (pos > 0 && nums2[pos] == nums2[pos - 1]) {
                pos--;
            }
            if (nums1[i] + nums2[pos] != pairSum) {
                continue;
            }
            int count = (int) min((long) k, (long) (i - start1 + 1) * (start2 - pos + 1));
            for (int j = 0; j < count && k > 0; j++, k--) {
                ans.push_back({nums1[i], nums2[pos]});
            }
        }
        return ans;
    }

    long min(long num1, long num2) {
        return num1 <= num2 ? num1 : num2;
    }
};