#leetcode#220. Contains Duplicate III
阿新 • • 發佈:2018-12-27
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
--------------------------------------
兩種解法, 1, bucket。 2, Treeset
相通之處在於都是維護一個長度為K的sliding window來確保 absolute difference between i and j is at most k.
轉換成long避免溢位, 比如 Integer.MIN_VALUE (-2147483648, -2147483647)k = 3, t = 3 這種input
Bucket:O(n)
public class Solution { public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) { if(k < 1 || t < 0) return false; long valueDiff = (long)t; long indexDiff = (long)k; Map<Long, Long> map = new HashMap<>(); for(int i = 0; i < nums.length; i++){ long value = (long)nums[i] - Integer.MIN_VALUE; long bucket = value / (valueDiff + 1); if(map.containsKey(bucket) || map.containsKey(bucket - 1) && value - map.get(bucket - 1) <= valueDiff || map.containsKey(bucket + 1) && map.get(bucket + 1) - value <= valueDiff) return true; map.put(bucket, value); if(i >= k){ long idx = ((long)nums[i - k] - Integer.MIN_VALUE) / (valueDiff + 1); map.remove(idx); } } return false; } }
TreeSet: O(nlogk)
public class Solution {
public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
TreeSet<Long> set = new TreeSet<>();
for(int i = 0; i < nums.length; i++){
Long ceil = set.ceiling((long)nums[i] - (long)t);
Long floor = set.floor((long)nums[i] + (long)t);
if(ceil != null && ceil <= nums[i] || floor != null && floor >= nums[i])
return true;
set.add((long)nums[i]);
if(i >= k)
set.remove((long)nums[i - k]);
}
return false;
}
}