373. 查找和最小的K对数字 :「多路归并」&「二分」

输入: nums1 = [1,7,11], nums2 = [2,4,6], k = 3
输出: [1,2],[1,4],[1,6]
解释: 返回序列中的前 3 对数：
     [1,2],[1,4],[1,6],[7,2],[7,4],[11,2],[7,6],[11,4],[11,6]
复制代码

输入: nums1 = [1,1,2], nums2 = [1,2,3], k = 2
输出: [1,1],[1,1]
解释: 返回序列中的前 2 对数：
     [1,1],[1,1],[1,2],[2,1],[1,2],[2,2],[1,3],[1,3],[2,3]
复制代码

输入: nums1 = [1,2], nums2 = [3], k = 3 
输出: [1,3],[2,3]
解释: 也可能序列中所有的数对都被返回:[1,3],[2,3]
复制代码

class Solution {
    boolean flag = true;
    public List<List<Integer>> kSmallestPairs(int[] nums1, int[] nums2, int k) {
        List<List<Integer>> ans = new ArrayList<>();
        int n = nums1.length, m = nums2.length;
        if (n > m && !(flag = false)) return kSmallestPairs(nums2, nums1, k);
        PriorityQueue<int[]> q = new PriorityQueue<>((a,b)->(nums1[a[0]]+nums2[a[1]])-(nums1[b[0]]+nums2[b[1]]));
        for (int i = 0; i < Math.min(n, k); i++) q.add(new int[]{i, 0});
        while (ans.size() < k && !q.isEmpty()) {
            int[] poll = q.poll();
            int a = poll[0], b = poll[1];
            ans.add(new ArrayList<>(){{
                add(flag ? nums1[a] : nums2[b]);
                add(flag ? nums2[b] : nums1[a]);
            }});
            if (b + 1 < m) q.add(new int[]{a, b + 1});
        }
        return ans;
    }
}
复制代码

class Solution:
    def kSmallestPairs(self, nums1: List[int], nums2: List[int], k: int) -> List[List[int]]:
        flag, ans = (n := len(nums1)) > (m := len(nums2)), []
        if flag:
            n, m, nums1, nums2 = m, n, nums2, nums1
        pq = []
        for i in range(min(n, k)):
            heapq.heappush(pq, (nums1[i] + nums2[0], i, 0))
        while len(ans) < k and pq:
            _, a, b = heapq.heappop(pq)
            ans.append([nums2[b], nums1[a]] if flag else [nums1[a], nums2[b]])
            if b + 1 < m:
                heapq.heappush(pq, (nums1[a] + nums2[b + 1], a, b + 1))
        return ans
复制代码

func kSmallestPairs(nums1 []int, nums2 []int, k int) [][]int {
    n, m, ans := len(nums1), len(nums2), [][]int{}
    flag := n > m
    if flag {
        n, m, nums1, nums2 = m, n, nums2, nums1
    }
    if n > k {
        n = k
    }
    pq := make(hp, n)
    for i := 0; i < n; i++ {
        pq[i] = []int{nums1[i] + nums2[0], i, 0}
    }
    heap.Init(&pq)
    for pq.Len() > 0 && len(ans) < k {
        poll := heap.Pop(&pq).([]int)
        a, b := poll[1], poll[2] 
        if flag{
            ans = append(ans, []int{nums2[b], nums1[a]})
        }else{
            ans = append(ans, []int{nums1[a], nums2[b]})
        }
        if b < m - 1 {
            heap.Push(&pq, []int{nums1[a] + nums2[b + 1], a, b + 1})
        }
    }
    return ans
}
// 最小堆模板
type hp [][]int
func (h hp) Len() int            { return len(h) }
func (h hp) Less(i, j int) bool  { return h[i][0] < h[j][0] }
func (h hp) Swap(i, j int)       { h[i], h[j] = h[j], h[i] }
func (h *hp) Push(v interface{}) { *h = append(*h, v.([]int)) }
func (h *hp) Pop() interface{}   { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }
复制代码

class Solution {
    int[] nums1, nums2;
    int n, m;
    public List<List<Integer>> kSmallestPairs(int[] n1, int[] n2, int k) {
        nums1 = n1; nums2 = n2;
        n = nums1.length; m = nums2.length;
        List<List<Integer>> ans = new ArrayList<>();
        int l = nums1[0] + nums2[0], r = nums1[n - 1] + nums2[m - 1];
        while (l < r) {
            int mid = l + r >> 1;
            if (check(mid, k)) r = mid;
            else l = mid + 1;
        }
        int x = r;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (nums1[i] + nums2[j] < x) {
                    List<Integer> temp = new ArrayList<>();
                    temp.add(nums1[i]); temp.add(nums2[j]);
                    ans.add(temp);
                } else break;
            }
        }
        for (int i = 0; i < n && ans.size() < k; i++) {
            int a = nums1[i], b = x - a;
            int c = -1, d = -1;
            l = 0; r = m - 1;
            while (l < r) {
                int mid = l + r >> 1;
                if (nums2[mid] >= b) r = mid;
                else l = mid + 1;
            }
            if (nums2[r] != b) continue;
            c = r;
            l = 0; r = m - 1;
            while (l < r) {
                int mid = l + r + 1 >> 1;
                if (nums2[mid] <= b) l = mid;
                else r = mid - 1;
            }
            d = r;
            for (int p = c; p <= d && ans.size() < k; p++) {
                List<Integer> temp = new ArrayList<>();
                temp.add(a); temp.add(b);
                ans.add(temp);
            }
        }
        return ans;
    }
    boolean check(int x, int k) {
        int ans = 0;
        for (int i = 0; i < n && ans < k; i++) {
            for (int j = 0; j < m && ans < k; j++) {
                if (nums1[i] + nums2[j] <= x) ans++;
                else break;
            }
        }
        return ans >= k;
    }
}
复制代码