哈希（3）

原题：

/** * Created by gouthamvidyapradhan on 28/03/2017. * 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. * <p> * 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]. */

答案：

public class KdiffPairsInanArray {    private Map<Integer, Integer> map = new HashMap<>();    private int count = 0;    /**     * Main method     *     * @param args     * @throws Exception     */    public static void main(String[] args) throws Exception {        int[] nums = {1, 2, 3, 4, 5};        System.out.println(new KdiffPairsInanArray().findPairs(nums, -1));    }    public int findPairs(int[] nums, int k) {        if (nums.length == 0 || k < 0) return 0;        for (int i : nums) {            map.put(i, map.getOrDefault(i, 0) + 1);        }        for (Map.Entry<Integer, Integer> entry : map.entrySet()) {            if (k == 0) {                if (entry.getValue() > 1)                    count++;            } else {                if (map.containsKey(entry.getKey() + k))                    count++;            }        }        return count;    }}