LeetCode 352. Data Stream as Disjoint Intervals

原题链接在这里：https://leetcode.com/problems/data-stream-as-disjoint-intervals/description/

题目：

Given a data stream input of non-negative integers a1, a2, ..., an, ..., summarize the numbers seen so far as a list of disjoint intervals.

For example, suppose the integers from the data stream are 1, 3, 7, 2, 6, ..., then the summary will be:

[1, 1]
[1, 1], [3, 3]
[1, 1], [3, 3], [7, 7]
[1, 3], [7, 7]
[1, 3], [6, 7]

Follow up:
What if there are lots of merges and the number of disjoint intervals are small compared to the data stream's size?

题解：

利用TreeMap<int[]> tm来保存每段Interval的start 和 Interval本身的对应关系. 

新值进来就看能不能连上已有的Interval.

Note:

1. If val is in tm, return. 

2. If val is within lower key corresponding interval, return.

Time Complexity: addNum, O(logn). getIntervals, O(n).

Space: O(n).

AC Java:

class SummaryRanges {
    TreeMap<Integer, int[]> tm;
    
    public SummaryRanges() {
        tm = new TreeMap<>();
    }
    
    public void addNum(int val) {
        if(tm.containsKey(val)){
            return;
        }
        
        Integer l = tm.lowerKey(val);
        Integer r = tm.higherKey(val);
        if(l != null && r != null && tm.get(l)[1] + 1 == val && val + 1 == r){
            tm.get(l)[1] = tm.get(r)[1];
            tm.remove(r);
        }else if(l != null && tm.get(l)[1] + 1 >= val){
            tm.get(l)[1] = Math.max(tm.get(l)[1], val);
        }else if(r != null && val + 1 == r){
            tm.put(val, new int[]{val, tm.get(r)[1]});
            tm.remove(r);
        }else{
            tm.put(val, new int[]{val, val});
        }
    }
    
    public int[][] getIntervals() {
        return tm.values().toArray(new int[0][0]);
    }
}

/**
 * Your SummaryRanges object will be instantiated and called as such:
 * SummaryRanges obj = new SummaryRanges();
 * obj.addNum(val);
 * int[][] param_2 = obj.getIntervals();
 */

类似Missing Ranges, Summary Ranges.