K-diff Pairs in an Array：统计数组中差值为K的元素对

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:

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].

思路：由于有重复元素，要考虑去重，因此需要排序。对排好序的元素以i、j判断差值大小，进而调整i、j位置。

class Solution {
	 public int findPairs(int[] nums, int k) {
		 int len = nums.length;
		 if(len <= 0) return 0;
		 Arrays.sort(nums);
		 int i = 0,j = 0;
		 int result = 0;
		 int preI = Integer.MIN_VALUE;
		 int preJ = Integer.MIN_VALUE;
		 while(i < len && j < len){
			 while(( i == j )||(preI == nums[i] )&&( preJ == nums[j])){				 
				 j++;
				 if(i >= len || j >= len) return result;
			 }
			 long temp = Math.abs((long)nums[i] - (long)nums[j]);//防止溢出
			 if(temp == k){
				 result ++;
				 preI = nums[i];
				 preJ = nums[j];
				 i++;
			 }else if(temp < k){				
				 preJ = nums[j];
				 j++;
			 }else{
				 preI = nums[i];				
				 i++;
			 }
		 }
		 return result;
	 }
}