leetcode 532. K-diff Pairs in an Array

532. 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].

1、绝对值不可能为负数。

2、k有可能为0，所以要先 it->second --;

3、map必须先判断在不在里面，再判断value是否大于0。不能直接判断，否则map会自动创建。

class Solution {
public:
int findPairs(vector<int>& nums, int k)
{
if (k < 0) return 0;
map<int, int> mp;
for (auto it : nums)
mp[it] ++;
int ret = 0;
for (auto it = mp.begin(); it != mp.end(); it++)
{
it->second --;
if (mp.find(it->first + k) != mp.end() && mp[it->first + k] > 0)
ret ++;
}
return ret;
}
};