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题目

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

我的想法

和II一样用双指针

//记录一下错误做法class Solution {
       public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
           if(k == 0 || nums.length <= 1) return false;        int i = 0, j = 0;        while(i < nums.length-1){
               if(j - i == k || j == nums.length-1) {
   j = ++i; continue;}            j++;            if(Math.abs(nums[j]-nums[i]) <= t) return true;        }//改为Math.abs((long)nums[j]-(long)nums[i]即可        return false;    }}

解答

leetcode solution 1: Using TreeSet

public class Solution {
       public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
           TreeSet
   
     set = new TreeSet<>();        for (int i = 0; i < nums.length; i++) {
               Long onLower = set.ceiling((long)nums[i] - (long)t);             if (onLower != null && onLower <=  ((long)nums[i] + (long)t)) return true;            set.add((long)nums[i]);            if (i >= k) set.remove((long)nums[i-k]);        }        return false;    }}
   

leetcode solution 2: Using HashMap

public class Solution {
       public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
           if (k < 1 || t < 0) return false;        Map
   
     map = new HashMap<>();        for (int i = 0; i < nums.length; i++) {
               long remappedNum = (long) nums[i] - Integer.MIN_VALUE;            long bucket = remappedNum / ((long) t + 1);            if (map.containsKey(bucket)                    || (map.containsKey(bucket - 1) && remappedNum - map.get(bucket - 1) <= t)                        || (map.containsKey(bucket + 1) && map.get(bucket + 1) - remappedNum <= t))                            return true;            if (map.entrySet().size() >= k) {
                   long lastBucket = ((long) nums[i - k] - Integer.MIN_VALUE) / ((long) t + 1);                map.remove(lastBucket);            }            map.put(bucket, remappedNum);        }        return false;    }}
   

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