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：
