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:
- 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].
题目的意思是给定一个值K，从数组中找出差值为k元素的个数。
public int findPairs(int[] nums, int k) {
if(nums == null || nums.length ==0 || k < 0) return 0;
HashMap<Integer,Integer> map = new HashMap<>();
int count = 0;
for(int i:nums)
{
map.put(i, map.getOrDefault(i, 0) + 1);
}
for(Map.Entry<Integer, Integer> entry : map.entrySet())
{
if(k == 0)
{
if(entry.getValue() >= 2)
count ++;
}
else {
if(map.containsKey(entry.getKey() +k))
count ++;
}
}
return count;
}
代码的思想很简单 ，先使用map计算每个值出现的次数，假设nums=[3, 1, 4, 1, 5] k=2，那么map的结果是[1=2, 3=1, 4=1, 5=1]
- 1.如果k=0，找出出现次数大于2元素的个数
- 2.如果k ！= 0，这个时候巧妙的使用set，找出是否存在
entry.getKey() +k的值。
本题目的要点是map和set的使用。