LeetCode刷题之K-diff Pairs in an Ar

Problem

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.

Example1

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.

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).

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].

My Solution

class Solution {
    public void quickSort(int[] nums, int left, int right) {
       if (left > right) {
            return ;
       }
       int i = left, j = right, base = nums[left], temp = 0;
       while (i != j) {
            while (nums[j] >= base && i < j) {
                --j;
            }
            while (nums[i] <= base && i < j) {
                ++i;
            }
            if (i < j) {
                temp = nums[i];
                nums[i] = nums[j];
                nums[j] = temp;
            }
        }
        nums[left] = nums[i];
        nums[i] = base;

        quickSort(nums, left, i - 1);
        quickSort(nums, i + 1, right);
        return ;
    }

    public int[] removeDuplicates(int[] nums) {
        int[] result = new int[2];
        if (nums.length == 0) result[0] = 0;
        int i = 0;
        int count = 0;
        for (int j = 1; j < nums.length; ++j) {
            if (nums[j] != nums[i]) {
                ++i;
                nums[i] = nums[j];
                if (count >= 1) {
                    result[1]++;
                }
                count = 0;
            } else {
                count++;
            }
        }
        if (count >= 1) {
            result[1]++;
        }
        result[0] = i + 1;
        return result;
    }

    public int findPairs(int[] nums, int k) {
        int count = 0, t = 0;
        quickSort(nums, 0, nums.length - 1);
        int[] result = removeDuplicates(nums);
        int len = result[0];
        if (k == 0) {
            return result[1];
        }
        for (int i = 0; i < len; ++i) {
            for (int j = i + 1; j < len; ++j) {
                if (Math.abs(nums[i] - nums[j]) == k) {
                    t++;
                }
            }
            if (t >= 2) {
                count += 2;
                t = 0;
            } else {
                count += t;
                t = 0;
            }
        }
        return count;
    }
}

Great Solution

public int findPairs(int[] nums, int k) {
    if (nums == null || nums.length == 0 || k < 0)   return 0;
    
    Map<Integer, Integer> map = new HashMap<>();
    int count = 0;
    for (int i : nums) {
        map.put(i, map.getOrDefault(i, 0) + 1);
    }
    
    for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
        if (k == 0) {
            //count how many elements in the array that appear more than twice.
            if (entry.getValue() >= 2) {
                count++;
            } 
        } else {
            if (map.containsKey(entry.getKey() + k)) {
                count++;
            }
        }
    }
    
    return count;
}