Leetcode 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.

解题思路：

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/