0532. K-diff Pairs in an Array (M)

K-diff Pairs in an Array (M)

题目

Given an array of integers nums and an integer k, return the number of unique k-diff pairs in the array.

A k-diff pair is an integer pair (nums[i], nums[j]), where the following are true:

• 0 <= i, j < nums.length
• i != j
• a <= b
• b - a == k

Example 1:

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

Example 4:

Input: nums = [1,2,4,4,3,3,0,9,2,3], k = 3
Output: 2

Example 5:

Input: nums = [-1,-2,-3], k = 1
Output: 2

Constraints:

• 1 <= nums.length <= 104
• -107 <= nums[i] <= 107
• 0 <= k <= 107

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题意

在数组中找到满足条件的不重复的数对对数。

思路

1. 先排序，再对于每一个不重复的数字，用二分法找对应的数；
2. 先存入Hash表，再找成对的数，注意k=0的情况。

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代码实现

Java

二分搜索

class Solution {
    public int findPairs(int[] nums, int k) {
        int ans = 0;
        Arrays.sort(nums);
        for (int i = 0; i < nums.length; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            ans += search(nums, i + 1, nums[i] + k) ? 1 : 0;
        }
        return ans;
    }

    private boolean search(int[] nums, int index, int target) {
        if (index == nums.length) {
            return false;
        }
        int left = index, right = nums.length - 1;
        while (left <= right) {
            int mid = (right - left) / 2 + left;
            if (nums[mid] < target) {
                left = mid + 1;
            } else if (nums[mid] > target) {
                right = mid - 1;
            } else {
                return true;
            }
        }
        return false;
    }
}

Hash

class Solution {
    public int findPairs(int[] nums, int k) {
        int ans = 0;
        Map<Integer, Integer> hash = new HashMap<>();
        for (int num : nums) {
            hash.put(num, hash.getOrDefault(num, 0) + 1);
        }
        for (int num : hash.keySet()) {
            ans += k == 0 ? hash.get(num) >= 2 ? 1 : 0 : hash.containsKey(num + k) ? 1 : 0;
        }
        return ans;
    }
}