LeetCode K-diff Pairs in an Array
原题链接在这里:https://leetcode.com/problems/k-diff-pairs-in-an-array/
题目:
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].
题解:
HashMap<Integer, Integer> hm 计数 num与出现次数.
再iterate一遍hm, 看是否key+k也在hm中.
Note: corner case 例如 k<0.
Time Complexity: O(n), n = nums.length. Space: O(n).
AC Java:
public class Solution {
public int findPairs(int[] nums, int k) {
if(nums == null || nums.length == 0 || k < 0){
return 0;
} int res = 0;
HashMap<Integer, Integer> hm = new HashMap<Integer, Integer>();
for(int num : nums){
hm.put(num, hm.getOrDefault(num, 0)+1);
} for(Map.Entry<Integer, Integer> entry : hm.entrySet()){
if(k == 0){
if(entry.getValue() > 1){
res++;
}
}else{
if(hm.containsKey(entry.getKey()+k)){
res++;
}
}
}
return res;
}
}
类似Two Sum.
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