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 is k.
Example 1:
Input: [3, 1, 4, 1, 5], k = 2Output: 2Explanation: 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 = 1Output: 4Explanation: 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 = 0Output: 1Explanation: 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].
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方法一
思路：每个pair(a,b)，其中a<=b，a-b=k。可以看到只要确定了a和k，就可以找到b，并且为了保证pair的唯一性，找到b以后，b不能再作为其他pair的右半部分。具体见代码。
代码：
1 public class Solution { 2 public int findPairs(int[] nums, int k) { 3 if(nums == null || nums.length == 0 || k < 0) return 0; 4 Map map = new HashMap (); 5 for(int i = 0; i < nums.length; ++i) { 6 map.put(nums[i], i); 7 } 8 int res = 0; 9 for(int i = 0; i < nums.length; ++i) {10 if(map.containsKey(nums[i] + k) && map.get(nums[i] + k) != i) {11 map.remove(nums[i] + k);12 res++;13 } 14 }15 return res;16 }17 }
方法二
思路：具体见代码。
代码：
1 public class Solution { 2 public int findPairs(int[] nums, int k) { 3 if(nums == null || nums.length == 0 || k < 0) return 0; 4 Map map = new HashMap (); 5 int res = 0; 6 for(int i : nums) { 7 map.put(i, map.getOrDefault(i, 0) + 1); 8 } 9 if(k == 0) {10 for (Map.Entry entry : map.entrySet()) {11 if(entry.getValue() >= 2) res++;12 } 13 }else {14 for (Map.Entry entry : map.entrySet()) {15 if(map.containsKey(entry.getKey() + k)) res++;16 }17 }18 return res;19 }20 }