Medium
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
思路:采用桶排序,內插補點在t範圍内的盡量放在同一個桶裡,盡管如此也可能存在相鄰的桶裡,是以,每次也要把相鄰的桶考慮進去;把視窗大小維持不超過k;
class Solution {
public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
if(nums==null || nums.length<2 || k<1 || t<0) return false;
Map<Long,Long> map = new HashMap<>();
for(int i=0;i<nums.length;i++){
long temp = (long)nums[i] - Integer.MIN_VALUE;//數組可能存在負數
long bucket = temp/((long)t+1);//防止t+1溢出
if(map.containsKey(bucket)||(map.containsKey(bucket-1)&&temp-map.get(bucket-1)<=t)||(map.containsKey(bucket+1)&&map.get(bucket+1)-temp<=t)) //如果是把(bucket,nums[i])加到map中的話,會有nums[i]-map.get(bucket-1),在這裡必須轉long型,否則溢出
return true;
if(map.entrySet().size()>=k){維持在加進元素後視窗大小不超過k
long invalid = ((long)nums[i-k] - Integer.MIN_VALUE)/((long)t+1);
map.remove(invalid);
}
map.put(bucket,temp);
}
return false;
}
}