(medium)LeetCode 220.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 difference between nums[i] and nums[j] is at most t and the difference between i and j is at most k.

思想借鉴：维持一个长度为k的window, 每次检查新的值是否与原来窗口中的所有值的差值有小于等于t的. 如果用两个for循环会超时O(nk). 使用treeset( backed by binary search tree) 的subSet函数,可以快速搜索. 复杂度为 O(n logk)

代码如下：

import java.util.SortedSet;
public class Solution {
    public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
        if(k<1 || t<0 ||nums==null ||nums.length<2) return false;
        SortedSet<Long>set=new TreeSet<>();
        int len=nums.length;
        for(int i=0;i<len;i++){
            SortedSet<Long>subSet=set.subSet((long)nums[i]-t,(long)nums[i]+t+1);
            if(!subSet.isEmpty()) return true;
            if(i>=k)
               set.remove((long)nums[i-k]);
            set.add((long)nums[i]);   
        }
        return false;
    }
}

　　运行结果：