[Leetcode]532. K-diff Pairs in an Array
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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]
思路:先把数组按小到大排序。设当前的数为 p,判断 p 后面的数减去 p是否有等于 k 的,如果有等于k的,就增加k-diff-pair然后break(因为不能重复统计),
eg :[1,2,3,3],k=2; 到下标为 2 时,已经满足条件,就不用进行了,否则下个数还是3,这样就重复统计了。
令p = p下一个数。如此重复,知道p到最后一个数。
此外,我还设置了一个前驱pre等于当前数p,等p递增后,判断p是否等于pre,如果等于,则continue;(这么做也是为了不重复统计)
eg:[1,1,2,3],k=2;
class Solution { public int findPairs(int[] nums, int k) { if(nums.length==1) return 0; //如果这有一个数,返回0; int kDiffPair = 0; //统计满足条件的且不重复的有多少对 int pre = 0;
Arrays.sort(nums); for (int i=0;i<nums.length-1;i++){ if (i!=0&&nums[i]==pre) continue; // 如果是数组里第一个数,则前面肯定不会有重复的; for (int j = i+1;j<nums.length;j++){ if (nums[j]-nums[i]==k){ kDiffPair++; break; }
} pre = nums[i]; //让pre等于当前的数,为了下一轮循环判断下个数是否等于pre } return kDiffPair; } }
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