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].

这里先对数组进行升序排序，再用双指针找到差是k的pair，同时要避免相同的pair出现。代码如下：

public class Solution {
public int findPairs(int[] nums, int k) {
Arrays.sort(nums);
int i = 0, j = 1, count = 0;
while (i < nums.length && j < nums.length) {
if (i == j || nums[j] - nums[i] < k) {
j ++;
} else if (nums[j] - nums[i] > k){
i ++;
} else {
i ++;
count ++;
while (i < nums.length && nums[i] == nums[i - 1]) i ++;
j = Math.max(j + 1, i + 1);
}
}
return count;
}
}

还有一种方法是用Hashmap，存储所有的数，遍历keyset，如果k=0，就找出现次数大于2的数；否则，找有没有key+k的数。代码如下：

public class Solution {
public int findPairs(int[] nums, int k) {
if (nums == null || nums.length == 0 || k < 0)   return 0;

Map<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) {
//count how many elements in the array that appear more than twice.
if (entry.getValue() >= 2) {
count++;
}
} else {
if (map.containsKey(entry.getKey() + k)) {
count++;
}
}
}

return count;
}
}