leetcode--532. K-diff Pairs in an Array
2018-03-08 14:33
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532. K-diff Pairs in an Array
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.题目大意就是给定你一个数组和整数k,输出数组中任意两数差值为k的数目。有一对算一对但是相同的 [2,3][3,2]只统计一次。
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).
题解,根据理解容易程度:
解法1:类似于冒泡排序,不过要去重。
先对数组进行排序,这样就只用考虑一边的情况,例如:[1,2,1],k=1。如果不排序,[1,2],[2,1]都要进行考虑,这个怎么进行排除将是一个很麻烦的事情。
排序了之后[1,1,2],那么我们只要考虑第一层是否重复了。还不理解,那就看代码。
class Solution { public int findPairs(int[] nums, int k) { int count = 0; Arrays.sort(nums);//可以替换为快速排序 for(int i = 0;i < nums.length - 1 ;i++){ for(int j = i + 1;j < nums.length;j++){ if(nums[j] - nums[i] == k ){//不知道能不能用Math.abs() count++; break; } } //上边代码已经统计了一次,所以相同的向后移位 while(i < nums.length-1 && nums[i] == nums[i+1]){ i++; } } return count; } }
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