Leetcode 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:

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].

标签 Two Pointers Array

类似题目 (E) Minimum Absolute Difference in BST

 

 

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：

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<Integer, Integer>();
        for(int i = 0; i < nums.length; ++i) {
            map.put(nums[i], i);
        }
        int res = 0;
        for(int i = 0; i < nums.length; ++i) {
            if(map.containsKey(nums[i] + k) && map.get(nums[i] + k) != i) {
                map.remove(nums[i] + k);
                res++;
            } 
        }
        return res;
    }
}

方法二

思路：具体见代码。

代码：

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<Integer, Integer>();
        int res = 0;
        for(int i : nums) {
            map.put(i, map.getOrDefault(i, 0) + 1);
        }
        if(k == 0) {
            for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
                if(entry.getValue() >= 2) res++;
            }    
        }else {
            for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
                if(map.containsKey(entry.getKey() + k)) res++;
            }
        }
        return res;
    }
}