leetcode532. K-diff Pairs in an Array

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本文介绍了一种有效的算法来解决寻找数组中特定差值的唯一数对的问题。通过使用哈希表，该方法能够在O(n)的时间复杂度内找到所有唯一的k-diff数对。

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].

解法

将数组中的每个元素按照出现的次数放在map中，key为每个元素，value为出现的次数，如果k=0，次数为2，总数+1。如果k>0，判断当前元素+k是否在map中存在，如果存在，总数+1。

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

        int ret = 0;
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        for (int num : nums) {
            if (map.containsKey(num)) {
                map.put(num, map.get(num) + 1);
            } else {
                map.put(num, 1);
            }
        }
        for (Integer key : map.keySet()) {
            if (k > 0) {
                if (map.containsKey(key + k)) {
                    ret++;
                }
            } else {
                if (map.get(key) >= 2) {
                    ret++;
                }
            }
        }

        return ret;
    }
}