有些hard在于算法难，思路难想；有些hard在于corner case多，这题就是后者。

这题跟56. Merge Intervals是类似的，而且这个是已经排好了顺序给你；但是难度并没有下降，我觉得这种题目就是把所有可能的case都cover到就能AC了，但是往往可能你在纸上画都不一定能画得全，这可能也就是hard的原因（我在纸上画了）。这次我试了一下，一共调试了5次，start和end用max/min比较分别占了一次。这种题我感觉看答案只会被牵着鼻子走，自己要覆盖到所有的corner case才行。另外，一道题如果耗时太长脑力必然会严重下降。所以尽量速战速决。

 public List<Interval> insert(List<Interval> intervals, Interval newInterval) {List<Interval> res = new ArrayList<>();//corner case1if (intervals.isEmpty()) {res.add(newInterval);return res;}while (!intervals.isEmpty()) {if (intervals.get(0).start < newInterval.start && intervals.get(0).end < newInterval.start) {res.add(intervals.remove(0));} else {break;}}if (intervals.isEmpty()) {res.add(newInterval);return res;}//这里开始的intervals list第一个元素的end一定比newInterval的start大//1. won't overlappingif (intervals.get(0).start > newInterval.end) {res.add(newInterval);res.addAll(intervals);return res;}//2. wrapped by the 1st intervalif (intervals.get(0).start <= newInterval.start && intervals.get(0).end >= newInterval.end) {res.addAll(intervals);return res;}//3. overlappingInterval head = intervals.get(0);if (head == null) return res;int start = Math.min(head.start, newInterval.start);int end;while (newInterval.end >= head.end && intervals.size() > 1) {intervals.remove(0);head = intervals.get(0);}if (newInterval.end < intervals.get(0).start) {end = newInterval.end;res.add(new Interval(start, end));res.addAll(intervals);return res;}if (newInterval.end >= intervals.get(0).start) {end = Math.max(intervals.remove(0).end, newInterval.end);res.add(new Interval(start, end));res.addAll(intervals);return res;}// should have covered all possible casesreturn null;}
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短的方法见： https://discuss.leetcode.com/topic/7808/short-and-straight-forward-java-solution