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].

本題是尋找數組中差為k的數對的個數

方法一

 1     public int findPairs(int[] nums, int k) {
 2        if (nums == null || nums.length == 0 || k < 0)   return 0;
 3         
 4         Map<Integer, Integer> map = new
 
 HashMap<>();
 5         int count = 0;
 6         for (int i : nums) {
 7             map.put(i, map.getOrDefault(i, 0) + 1);
 8         }
 9         
10         for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
11             if (k == 0) {
12                 //count how many elements in the array that appear more than twice.
13                 if (entry.getValue() >= 2) {
14                     count++;
15                 } 
16             } else {
17                 if (map.containsKey(entry.getKey() + k)) {
18                     count++;
19                 }
20             }
21         }
22         
23         return count;
24     }

方法二

 1     public  int findPairs1(int[] nums, int k) {
 2         if(k<0 || nums.length<=1){
 3             return 0;
 4         }
 5 
 6         Arrays.sort(nums);
 7         int count = 0;
 8         int left = 0;
 9         int right = 1;
10 
11         while(right<nums.length){
12             int firNum = nums[left];
13             int secNum = nums[right];
14             if(secNum-firNum<k)
15                 right++;
16             else if(secNum - firNum>k)
17                 left++;
18             else{
19                 count++;
20                 while(left<nums.length && nums[left]==firNum){
21                     left++;
22                 }
23                 while(right<nums.length && nums[right]==secNum){
24                     right++;
25                 }
26 
27             }
28             if(right==left){
29                 right++;
30             }
31         }
32         return count;
33     }

532 K-diff Pairs in an Array