2017.11.22 LeetCode - 532. K-diff Pairs in an Array【简单二分】
2017-11-22 22:09
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532. K-diff Pairs in an Array
Description
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: 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 = 1Output: 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 = 0Output: 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].
题意: 给你一个数组,让你找出几对数(i,j),保证abs(i-j) == k
分析: 排完序,直接对每个数字二分下即可,(其实暴力也可以过,范围小),用二分的话,得考虑一下当k为0的情况,所以我在这里用了upper_bound,来看下右界,就可以方便的判重了
参考函数
class Solution { public: int findPairs(vector<int>& nums, int k) { int res = 0; sort(nums.begin(),nums.end()); for(int i = 0;i < nums.size();i++) { int t = upper_bound(nums.begin(),nums.end(),nums[i]+k)-nums.begin()-1; if(nums[t] - k == nums[i] && t > i) res++; while(nums[i] == nums[i+1]) i++; } return res; } };
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