532. k

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:

1
2
3

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:

1
2
3

Input: [1, 3, 1, 5, 4], k = 0
Output: 1
Explanation: There is one 0-diff pair in the array, (1, 1).

class  {
    public int findPairs(int[] nums, int k) {
        if (nums == null || nums.length == 0 || k < 0)   return 0;
        int count = 0;
      Map<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
         if (map.containsKey(nums[i])) {
             map.put(nums[i], map.get(nums[i]) + 1);
          } else {
                map.put(nums[i], 1);
         }
        }
        if (k != 0) {
           for (int i = 0; i < nums.length; i++) {
             if (map.containsKey(nums[i] - k)) {
                 count++;
                 map.remove(nums[i] - k);
             }
            }
        }
        if (k == 0) {
           Set<Integer> keySet = map.keySet();
            for (Integer key: keySet) {
             if (map.get(key) > 1) count++;
           }
        }
        return count;
    }
}

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