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].
本题方法是将数组值用hashmap来存起来,value来存储个数。如果k=0的话,value>=2的时候count+1,如果k不等于0,则看key+k的值在map里面是否包含,包含的话就count+1,代码如下:
1 public class Solution { 2 public int findPairs(int[] nums, int k) { 3 Map<Integer,Integer> map = new HashMap<Integer,Integer>(); 4 if(nums==null||nums.length==0||k<0) return 0; 5 for(int num:nums){ 6 map.put(num,map.getOrDefault(num,0)+1); 7 } 8 int count = 0; 9 for(Map.Entry<Integer,Integer> entry:map.entrySet()){ 10 if(k==0){ 11 if(entry.getValue()>=2){ 12 count++; 13 } 14 }else{ 15 if(map.containsKey(entry.getKey()+k)){ 16 count++; 17 } 18 } 19 } 20 return count; 21 } 22 }