K-diff Pairs in an Array

Given an array of integers and an integer k, you need to find the number of unique k-diff pairs in the array. Here a k-diff pair is defined as an integer pair (i, j), where i and j are both numbers in the array and their absolute difference is k.

Example 1:

Input: [3, 1, 4, 1, 5], k = 2

Output: 2

Explanation: There are two 2-diff pairs in the array, (1, 3) and (3, 5).

Although we have two 1s in the input, we should only return the number of unique pairs.

Example 2:

Input:[1, 2, 3, 4, 5], k = 1

Output: 4

Explanation: There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and (4, 5).

Example 3:

Input: [1, 3, 1, 5, 4], k = 0

Output: 1

Explanation: There is one 0-diff pair in the array, (1, 1).

Note:

The pairs (i, j) and (j, i) count as the same pair.

The length of the array won’t exceed 10,000.

All the integers in the given input belong to the range: [-1e7, 1e7].

方法1：排序+二分查找

class Solution {
private:
int binarySort(vector<int>& nums,int left,int right,int target){
int res = -1;
while(left<=right){
int middle = (left + right) / 2;
if(nums[middle]<target)
left = middle + 1;
else if(nums[middle] > target)
right = middle - 1;
else{
res = middle;
break;
}
}
return res;
}
public:
int findPairs(vector<int>& nums, int k) {
sort(nums.begin(),nums.end());
int count = 0;
for(int i = 0 ; i < nums.size() ; ++i){
if(i!=0&&nums[i]==nums[i-1])
continue;
int remain = k + nums[i];
if(binarySort(nums,i+1,nums.size()-1,remain)!=-1)
count++;
}
return count;
}
};

方法2：unordered_map

class Solution {

public:
int findPairs(vector<int>& nums, int k) {
if(k<0)
return 0;

unordered_map<int,int> mp;
for(auto num:nums){
++mp[num];
}
int count = 0;
if(k>0){
for(auto it = mp.begin();it!=mp.end();++it){
int key = it->first + k;
if(mp.find( key)!=mp.end())
++count;
}
}
else{
for(auto it = mp.begin();it!=mp.end();++it){
if(it->second>1)
++count;
}
}

return count;
}
};