Leetcode220

Contains Duplicate III:
Given an array of integers, find out whether there are two distinct indices i and j in the array such that the absolute difference between nums[i] and nums[j] is at most t and the absolute difference between i and j is at most k.
Example 1:
Input: nums = [1,2,3,1], k = 3, t = 0
Output: true
Example 2:
Input: nums = [1,0,1,1], k = 1, t = 2
Output: true
Example 3:
Input: nums = [1,5,9,1,5,9], k = 2, t = 3
Output: false

Solution:

#include 
#include 
#include 

using namespace std;

// 时间复杂度：o（nlogn）
// 空间复杂度：o（k）
class Solution {
public:
    bool containsNearbyAlmostDuplicate(vector<int>& nums, int k, int t) {

        set<long long> record;
        for( int i = 0; i < nums.size(); i++ ) {

            if (record.lower_bound( (long long)nums[i] - (long long)t) != record.end() &&
                *record.lower_bound( (long long)nums[i] - (long long)t) <= (long long)nums[i] + (long long)t)
                return true;

            record.insert(nums[i]);

            if (record.size() == k + 1)
                record.erase(nums[i - k]);
        }

        return false;

    }
};

总结： 因为set是用二叉树来构建的，所以时间复杂度为：o(nlogn)。lower_bound（）的意思是：返回大于等于val值的最小值的位置