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:
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].
Solution 1: sort,然后two pointers,注意去重
public int findPairs(int[] nums, int k) {
if (nums == null || nums.length < 2 || k < 0) {
return 0;
}
Arrays.sort(nums);
int i = 0, j = 1, ans = 0;
while (i < nums.length && j < nums.length) {
while (i - 1 >= 0 && nums[i] == nums[i - 1] && i < j - 1) { //是i < j-1,不是i<j,否则[1,1,3,4,5]出错
i++;
}
while (j + 1 <nums.length && nums[j] == nums[j + 1]) {
j++;
}
if (nums[j] - nums[i] == k) {
ans++;
i++;
j++;
} else if (nums[j] - nums[i] < k) {
j++;
} else {
i++;
}
if (i == j) {
j++;
}
}
return ans;
}
Solution 2: hashmap。注意k == 0的话,要有元素出现2次
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;
}