X和Y都是从大到小排好序的一维数组，数都是正数，数组size分别是m和n。他们一共可以组成m*n 个数对。求前K个和最大的数对。

解法:

heap.add(pair(0, 0));  // largest pair

// remove max k-1 times
for (int i = 0; i < k - 1; ++i) {
    // get max and remove it from the heap
    max = heap.popMax();

    // add next candidates
    heap.add(pair(max.i + 1, max.j));
    heap.add(pair(max.i, max.j + 1));
}

// get k-th maximum element
max = heap.popMax();
maxVal = a[max.i] + b[max.j];

Some things to consider.

- Duplicated pairs can be added to the heap, this can be prevented with hash.
- Indexes need to be validated, e.g. that max.i + 1 < a.length.


时间复杂度: O(klog(k))
空间复杂度: O(k)

参考资料:
[1] http://stackoverflow.com/questions/5212037/find-the-pair-across-2-arrays-with-kth-largest-sum