标签：
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.
关键是要想到用TreeSet或者SortedSet, 而且要维护好k 的窗口！！！
另外，要考虑溢出情况，所以必须使用long !!!!
方法一： TreeSet 用的是Binary Search Tree, 有两个有用的method: ceiling() and floor(), the time complexity is O(nlog(k)).
ceiling() : Returns the least element in this set greater than or equal to the given element, or null if there is no such element.
floor() : Returns the greatest element in this set less than or equal to the given element, or null if there is no such element.
方法二： SortedSet 可以界定左边界和右边界。
两个方法的共同点是维护好k的窗口。
Java code:
方法一：
public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) { if(k < 1 || t < 0) { return false; } TreeSet<Long> set = new TreeSet<Long>(); for(int i= 0; i< nums.length; i++) { long x = (long)nums[i]; if((set.floor(x) != null && (x-(long)(set.floor(x)) <=(long)t)) || (set.ceiling(x)!=null && ((long)(set.ceiling(x))-x <=(long)t))) { return true; } set.add(x); if(i>=k) { set.remove((long)nums[i-k]); } } return false; }
方法二：
public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) { if(k < 1 || t < 0) { return false; } SortedSet<Long> set = new TreeSet<Long>(); for(int i =0; i< nums.length; i++) { long leftBound = (long)nums[i]-t; long rightBound = (long)nums[i]+t+1; SortedSet<Long> subSet = set.subSet(leftBound, rightBound); if(!subSet.isEmpty()){ return true; } set.add((long)nums[i]); if(i>=k) { set.remove((long)nums[i-k]); } } return false; }
Reference:
1. http://www.programcreek.com/2014/06/leetcode-contains-duplicate-iii-java/
Leetcode Contains Duplicate III
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原文地址：http://www.cnblogs.com/anne-vista/p/4827824.html
