blueWind123731
commented
3 years ago 双指针
将nums排序后，对应一个前指针i，移动后指针j查看是否有 nums[i]+k==nums[j]。如果有，则计数器加1.
为了避免重复，保证前指针i跳过重复的元素。
另外注意的一个细节就是，每确定一个指针i，都要重新定位j=i+1.
var findPairs = function(nums, k) {
if(nums.length<2)return 0
nums.sort((a,b)=>a-b)
let n = nums.length
let i=0,j=1
let count = 0
while(j<n){
j = i+1
while(j<n&&nums[j]<nums[i]+k)j++
if(j<n&&nums[j]==nums[i]+k){
count++
}
i++
while(i<n&&nums[i]===nums[i-1])i++
}
return count
};
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 |nums[i] - nums[j]| == k Notice that |val| denotes the absolute value of val.
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
https://leetcode.com/problems/k-diff-pairs-in-an-array/