LeetCode 532. K-diff Pairs in an Array (在数组中相差k的配对)
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:
- 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].
题目标签:Array, Two Pointers
题目给了我们一个nums array 和一个 k, 让我们找出有几对 相差k 的配对。
建立一个HashMap 把nums 的数字num 作为key 存入;把数字num 出现的 次数当作value 存入。
遍历map 的keys, 找 key + k 在map 里存不存在,存在的话说明这一对是k-diff pair。这种情况是当k 不等于 0;
当k = 0, 就要用到map 里的value 了,当key 出现的次数大于1 的时候, 至少2,那么 key 和 key 也是一对k-diff pair。
Java Solution:
Runtime beats 26.84%
完成日期:05/10/2017
关键词:Array, HashMap
关键点:把num当作key,把出现次数当作value 存入map,再遍历keys 来找k-diff pair
1 public class Solution 2 { 3 public int findPairs(int[] nums, int k) 4 { 5 int res = 0; 6 HashMap<Integer, Integer> map = new HashMap<>(); 7 8 // set up the HashMap, key: number; value: occurrence. 9 for(int i=0; i<nums.length; i++) 10 { 11 // if map doesn't have it, add it into map and set value to 1. 12 // if map has it, increase it value by 1. 13 map.put(nums[i], map.getOrDefault(nums[i], 0) + 1); 14 15 } 16 17 // iterate the map 18 for(Integer key: map.keySet()) 19 { 20 if(k == 0 && map.get(key) > 1) // if k == 0 , only check it's value 21 res++; 22 else if(k > 0 && map.containsKey(key + k)) // if k > 0, check the map has (key + k) as a key 23 res++; 24 25 } 26 27 return res; 28 } 29 }
参考资料:
http://www.cnblogs.com/grandyang/p/6545075.html
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