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
题意
在数组中找到满足条件的不重复的数对对数。
思路
- 先排序,再对于每一个不重复的数字,用二分法找对应的数;
- 先存入Hash表,再找成对的数,注意k=0的情况。
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;
}
}