simple implementation of a set... sorted list with no duplicates
O(log n) to see if contains an item (binary search)
O(n) to add an item in sorted position

interesting methods: union, intersection, subtract... all variants of the 
merge routine that is part of mergesort

union:
while(ia<a.length && ib<b.length) {
  if (a[ia]<b[ib]) {
    result[k++]=a[ia++];
  } else if (a[ia]>b[ib]) {
    result[k++]=b[ib++];    
  } else { // ==
    result[k++]=b[ib++];    
    ia++;
  }
}
while(ia<a.length)
  result[k++]=a[ia++];
while(ib<b.length)
  result[k++]=b[ib++];

intersection:
while(ia<a.length && ib<b.length) {
  if (a[ia]<b[ib]) {
    ia++;
  } else if (a[ia]>b[ib]) {
    ib++;
  } else { // ==
    result[k++]=b[ib++];    
    ia++;
  }
}

subtract:
while(ia<a.length && ib<b.length) {
  if (a[ia]<b[ib]) {
    result[k++]=a[ia++];
  } else if (a[ia]>b[ib]) {
    ib++;
  } else { // ==
    ia++; ib++;
  }
}
while(ia<a.length)
  result[k++]=a[ia++];


TreeMap is ordered, ops are O(log n).  because HashMap is not ordered,
ops can magically get down to O(1).


a set can be thought of as a map where the values can only be true

Map<Key,Value>

power of HashMap:

given an array of ints, find if any two sum to k

for(int i=0;i<arr.length;i++) {
  if (hashset.contains(arr[i})) return true;
  hashset.add(k-arr[i]);
}
return false;