leetcode 719. Find K-th Smallest Pair Distance

719. Find K-th Smallest Pair Distance

Given an integer array, return the k-th smallest distance among all the pairs. The distance of a pair (A, B) is defined as the absolute difference between A and B. 

Example 1:

Input:
nums = [1,3,1]
k = 1
Output: 0
Explanation:
Here are all the pairs:
(1,3) -> 2
(1,1) -> 0
(3,1) -> 2
Then the 1st smallest distance pair is (1,1), and its distance is 0.

Note:

2 <= len(nums) <= 10000

.

0 <= nums[i] < 1000000

.

1 <= k <= len(nums) * (len(nums) - 1) / 2

.

最开始想的方法。 TLE。

class Solution {
public:
int smallestDistancePair(vector<int>& nums, int k)
{
map<int, int> mmp;
map<int, int> num;
for (int i = 0; i < nums.size(); i++)
{
num[nums[i]]++;
}

for (auto it = num.begin(); it != num.end(); it++)
{
//还有0需要填入
if (it->second >= 2)
mmp[0] += it->second * (it->second - 1) / 2;

auto itt = it;
itt++;
for (; itt != num.end(); itt++)
{
mmp[abs(it->first - itt->first)] += it->second * itt->second;
}
}

for (auto it = mmp.begin(); it != mmp.end(); it ++)
{
if (k > it->second)
{
k = k - it->second;
continue;
}
else
{
return it->first;
}
}
return 0;
}
};

然后去网上找，发现可以用二分法 + 滑动窗口。
1、二分答案。缩小范围。

2、滑动窗口可以减少运算量。

class Solution {
public:
int smallestDistancePair(vector<int>& nums, int k)
{
//二分答案
sort(nums.begin(), nums.end());
int start = 0;
int end = nums[nums.size() - 1] - nums[0];
while (start + 1 < end)
{
int mid = (end - start) / 2 + start;
if (count(nums, mid) >= k)//求的是 nums之间的差值 小于等于 mid 的个数。
end = mid;
else
start = mid;
}
//double check
if ( count(nums, start) >= k)
return start;
else
return end;
}

int count(vector<int>& nums, int mid) //差值 小于等于 mid 的个数
{
//search
//使用窗口思想，判断差值<=k的个数，r-l即表示[l,r]间间隔<m的个数（每确定一个窗口就新增加了（r-l+1）- 1个差值对）
int left = 0;
int ret = 0;
for(int right = 0; right < nums.size(); right++)
{
while (nums[right] - nums[left] > mid)
{
left++;
}
ret += right - left;
}
return ret;
}
};