[LeetCode] 373. Find K Pairs with Smallest Sums @ python

一．题目：
从两个数组中各拿出一个数字，求两个数字的和最小的k个组合。如果和相等的时候，出现的顺序是不在乎的。
Example 1:
Input: nums1 = [1,7,11], nums2 = [2,4,6], k = 3
Output: [[1,2],[1,4],[1,6]]
Explanation: The first 3 pairs are returned from the sequence:
[1,2],[1,4],[1,6],[7,2],[7,4],[11,2],[7,6],[11,4],[11,6]
Example 2:
Input: nums1 = [1,1,2], nums2 = [1,2,3], k = 2
Output: [1,1],[1,1]
Explanation: The first 2 pairs are returned from the sequence:
[1,1],[1,1],[1,2],[2,1],[1,2],[2,2],[1,3],[1,3],[2,3]
Example 3:
Input: nums1 = [1,2], nums2 = [3], k = 3
Output: [1,3],[2,3]
Explanation: All possible pairs are returned from the sequence: [1,3],[2,3］

二．解题思路：
（１）方法一
从例１中我们可以得到暴力的解法，首先递归得到所有的组合，然后取出其中和最小的前k个组合．代码如下：

class Solution(object):
    def kSmallestPairs(self, nums1, nums2, k):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :type k: int
        :rtype: List[List[int]]
        """
        dicts = collections.defaultdict(list)
        lists = [nums1,nums2]
        def dfs(index,lists,res,dicts):
            if index == 2:
                dicts[sum(res)] +=[res]
                return
            for i in lists[index]:
                dfs(index+1,lists,res+[i],dicts)
        dfs(0,lists,[],dicts)
        result = []
        for key in sorted(dicts.keys()):
            for i in dicts[key]:
                if len(result) < k:
                    result.append(i)
                elif len(result) == k:
                    return result
        return result

（２）方法二
从各个组合中找到前k个和最小的组合，从这里看我们可以尝试使用堆来实现．首先将（nums1[i] + nums2[0], i, 0）加入堆，i取值范围[0, size1)；然后弹出堆顶元素sum, i, j，将(nums1[i], nums2[j])加入结果集ans，若j + 1 < size2，则将(nums1[i] + nums2[j + 1], i, j + 1)加入堆．代码如下：

class Solution(object):
    def kSmallestPairs(self, nums1, nums2, k):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :type k: int
        :rtype: List[List[int]]
        """
        res = []
        len1, len2 = len(nums1), len(nums2)
        if not len1 or not len2: return res
        heap = []
        for x in range(len1):
            heapq.heappush(heap, (nums1[x] + nums2[0], x, 0))
        while len(res) < min(k, len1 * len2):
            sum, i, j = heapq.heappop(heap)
            res.append([nums1[i], nums2[j]])
            if j + 1 < len2:
                heapq.heappush(heap, (nums1[i] + nums2[j + 1], i, j + 1))
        return res