[leetcode] 532. K-diff Pairs in an Array

Question:

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 = 2Output: 2Explanation: 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 = 1Output: 4Explanation: 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 = 0Output: 1Explanation: 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].

Solution:

分情况讨论：
1. k小于0。因为绝对值不会小于0，直接返回0。
2. k等于0。只有两个或以上相同的元素才算一对，因此还需一个visited数组来记录已访问过的第二个相同元素。
3. k大于0，对每一个元素，查看之前发现过的元素，看是否能有满足绝对值等于k的。

class Solution {public:    int findPairs(vector<int>& nums, int k) {        if (k < 0) return 0;        set<int> s;        int ret = 0;        if (k == 0) {            set<int> visited;            for (int i = 0; i < nums.size(); i++) {                if (visited.count(nums[i]) == 0 && s.count(nums[i]) > 0) {                    ret++;                    visited.insert(nums[i]);                }                s.insert(nums[i]);            }            return ret;        }        for (int i = 0; i < nums.size(); i++) {            if (s.count(nums[i]) > 0)                continue;            if (s.count(nums[i] - k) > 0)                ret++;            if (s.count(nums[i] + k) > 0)                ret++;            s.insert(nums[i]);        }        return ret;    }};