57 Insert Interval

题目链接：https://leetcode.com/problems/insert-interval/

题目：

Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).

You may assume that the intervals were initially sorted according to their start times.

Example 1:
Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].

Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].

This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].

解题思路：
解题前先理清可能发生的情况很重要。就本题来说，有三种情况。
要插入的区间在区间列表中
1、某个区间的内部
2、所有区间外
3、跨区间
其中，跨区间又分三种情况：

代码实现：

/**
 * 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 insert(List intervals, Interval newInterval) {
        List res = new ArrayList();
        if(newInterval == null)
            return intervals;
        if(newInterval != null && (intervals == null || intervals.size() == 0)) {
            res.add(newInterval);
            return res;
        }
        int start = -1;
        boolean flag = false;
        for(Interval i : intervals) {
            if(flag) { // 给定的区间已经被归并，不再与剩余区间做比较
                res.add(i);
                continue;
            }
            if(i.start <= newInterval.start && i.end >= newInterval.end) { // 区间内
                res.add(i);
                flag = true;
                continue;
            } else if(i.end < newInterval.start || i.start > newInterval.end) { // 区间外
                if(i.start > newInterval.end) {
                    if(start != -1)
                        res.add(new Interval(start, newInterval.end));
                    else res.add(newInterval);
                    flag = true;
                }
                res.add(i);
                continue;
            } else { // 跨区间
                if(start == -1) // 遇到第一个被跨区间的区间
                    start = Math.min(i.start, newInterval.start);
                if(i.end >= newInterval.end) {
                    res.add(new Interval(start, i.end));
                    flag = true;
                }
            }
        }
        if(!flag) {
            if(start != -1)
                res.add(new Interval(start, newInterval.end));
            else res.add(newInterval);
        }
        return res;
    }
}

151 / 151 test cases passed.
Status: Accepted
Runtime: 4 ms