## 题目描述：

LeetCode 373. Find K Pairs with Smallest Sums

You are given two integer arrays nums1 and nums2 sorted in ascending order and an integer k.

Define a pair (u,v) which consists of one element from the first array and one element from the second array.

Find the k pairs (u1,v1),(u2,v2) ...(uk,vk) with the smallest sums.

Example 1:

```Given nums1 = [1,7,11], nums2 = [2,4,6],  k = 3

Return: [1,2],[1,4],[1,6]

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:

```Given nums1 = [1,1,2], nums2 = [1,2,3],  k = 2

Return: [1,1],[1,1]

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:

```Given nums1 = [1,2], nums2 = [3],  k = 3

Return: [1,3],[2,3]

All possible pairs are returned from the sequence:
[1,3],[2,3]
```

## 解题思路：

```记nums1的下标为i，nums2下标为j；数组长度为size1，size2；

## Python代码：

``````class Solution(object):
def kSmallestPairs(self, nums1, nums2, k):
"""
:type nums1: List[int]
:type nums2: List[int]
:type k: int
:rtype: List[List[int]]
"""
ans = []
heap = [(0x7FFFFFFF, None, None)]
size1, size2 = len(nums1), len(nums2)
idx2 = 0
while len(ans) < min(k, size1 * size2):
if idx2 < size2:
sum, i, j = heap[0]
if nums2[idx2] + nums1[0] < sum:
for idx1 in range(size1):
heapq.heappush(heap, (nums1[idx1] + nums2[idx2], idx1, idx2))
idx2 += 1
sum, i, j = heapq.heappop(heap)
ans.append((nums1[i], nums2[j]))
return ans
``````

```首先将（nums1[i] + nums2[0], i, 0）加入堆，i取值范围[0, size1)

## Python代码：

``````class Solution(object):
def kSmallestPairs(self, nums1, nums2, k):
"""
:type nums1: List[int]
:type nums2: List[int]
:type k: int
:rtype: List[List[int]]
"""
ans = []
size1, size2 = len(nums1), len(nums2)
if size1 * size2 == 0: return ans
heap = []
for x in range(size1):
heapq.heappush(heap, (nums1[x] + nums2[0], x, 0))
while len(ans) < min(k, size1 * size2):
sum, i, j = heapq.heappop(heap)
ans.append((nums1[i], nums2[j]))
if j + 1 < size2:
heapq.heappush(heap, (nums1[i] + nums2[j + 1], i, j + 1))
return ans
``````

## Python代码：

``````def kSmallestPairs(self, nums1, nums2, k):
queue = []
def push(i, j):
if i < len(nums1) and j < len(nums2):
heapq.heappush(queue, [nums1[i] + nums2[j], i, j])
push(0, 0)
pairs = []
while queue and len(pairs) < k:
_, i, j = heapq.heappop(queue)
pairs.append([nums1[i], nums2[j]])
push(i, j + 1)
if j == 0:
push(i + 1, 0)
return pairs
``````

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