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]
题目解析:找出两个有序数组里面元素组合和最小的k对数。
这里,需要注意的是是边界问题。
方法很多,最小堆,优先队列等都可以,这里提供另一种思路:用一个数组存储每个nums1对应元素在nums2中组合索引
源代码:
class Solution {
public:
vector<pair<int, int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
vector<pair<int, int> > res;
int m = nums1.size();
int n = nums2.size();
int cnt = min(k, m * n);
vector<int> idx(m, 0);
while(cnt > 0)
{
int minSum = INT_MAX;
int curIdx = 0;
for (int i = 0; i < m; ++i) {
if (idx[i] < n && nums1[i] + nums2[idx[i]] <= minSum) {
minSum = nums1[i] + nums2[idx[i]];
curIdx = i;
}
}
res.push_back(make_pair(nums1[curIdx], nums2[idx[curIdx]]));
idx[curIdx]++;
cnt--;
}
return res;
}
};