這道題其實(shí)本身并不難。幾種情況分類討論有點(diǎn)亂。
沒寫出來，看了別人的答案：

/**
 * Definition for an interval.
 * public class Interval {
 *     int start;
 *     int end;
 *     Interval() { start = 0; end = 0; }
 *     Interval(int s, int e) { start = s; end = e; }
 * }
 */
public class Solution {
    public List<Interval> merge(List<Interval> intervals) {
        List<Interval> ans = new ArrayList<>();
        if (intervals == null || intervals.size()==0)   return ans;
        
        ans.add(intervals.get(0));
        for (int i=1; i<intervals.size(); i++) {
            Interval curr = intervals.get(i);
            int j = ans.size()-1;
            for (; j>=0; j--) {
                Interval prev = ans.get(j);
                if (prev.end < curr.start) {
                    ans.add(j+1, curr);
                    break;
                }
                else if (prev.start <= curr.start) {
                    prev.end = Math.max(prev.end, curr.end);
                    break;
                } 
                else if (prev.start <= curr.end) {
                    prev.start = curr.start;
                    prev.end = Math.max(prev.end, curr.end);
                    curr = prev;
                    ans.remove(j);
                }
                else {
                    continue;
                }
            }
            if (j<0)    ans.add(0,curr);
        }
        return ans;
    }
}