描述
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.
1.The pairs (i, j) and (j, i) count as the same pair.
2.The length of the array won't exceed 10,000.
3.All the integers in the given input belong to the range: [-1e7, 1e7].
您在真实的面试中是否遇到过这个题？
是
样例
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).这道题没有什么难度，用可以排序的关联容器就比较容易解决，但我开始用了set，我本来认为只要出现过就可以不必计数，但在第二个数据就没有过。
输入
[1,3,1,5,4]
0
输出
4
期望答案
1参考答案来看只有[1,1]才是正确的。
于是改用map容器，此时我又犯了一个错误，map容器中count函数只能判断是否存在，count的返回值只有0或者1.
class Solution {
public:
/**
* @param nums: an array of integers
* @param k: an integer
* @return: the number of unique k-diff pairs
*/
int findPairs(vector<int> &nums, int k) {
// Write your code here
map<int,int> m_map;
int sum=0;
for(int i=0;i<nums.size();i++)
m_map[nums[i]]++;
for (map<int, int> ::iterator it = m_map.begin(); it != m_map.end();it++) {
it->second--;
if (m_map.count(it->first + k)>0&&m_map[it->first + k]>0)
sum++;
}
return sum;
}
};