LeetCode -- Insert Interval

题目描述：




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].




本题要求，对于一个已排序好的间隔序列，插入一个新的间隔，需要保证已排序，并且合并overlap的间隔。




本题的复杂度在于情况数间隔的起始和结束位置的判断，需要分情况讨论。


思路：
1. 首先判断新间隔i的end是否比intervals[0].start小，如果是：插入在第一个。
2. 判断i.start是否比intervals.[Count].end 大，如果是：放最后
3. 遍历Intervals，对于start：
a.gap:start 位于intervals[i].end和intervals[i+1].start之间，startIndex = i
b.between:start 位于intervals[i].start,和intervals[i].end之间, startIndex = i-1;
c.start比intervals[0].start小,startIndex=-1
d.确定startIndex (之后用于确定是否添加到result中)


确定start的情况后，起1个内循环j∈[i,Count)找到end符合的情况：
a.gap: end 位于intervals[i].end和intervals[i+1].start之间 endIndex = j + 1
b.start 位于intervals[i].start,和intervals[i].end之间, endIndex = j+1
c.end比intervals[Count].end大, endIndex = Count+1
d.确定endIndex


4.1 : 将[0,startIndex]间的间隔添加到集合
4.2 : 将combine后的interval，添加到集合。
4.3 : 将[endIndex,Count)间的间隔添加到集合中




实现代码：

/** * Definition for an interval. * public class Interval { *     public int start; *     public int end; *     public Interval() { start = 0; end = 0; } *     public Interval(int s, int e) { start = s; end = e; } * } */public class Solution {    public IList<Interval> Insert(IList<Interval> intervals, Interval newInterval) {    // intervals is empty , return new interval        if(intervals.Count == 0){return new List<Interval>(){newInterval};}var result = new List<Interval>();if(newInterval.end < intervals[0].start){result.Add(newInterval);result.AddRange(intervals);return result;}if(newInterval.start > intervals[intervals.Count-1].end){result.AddRange(intervals);result.Add(newInterval);return result;}var newAsStart = false;if(newInterval.start < intervals[0].start){newAsStart = true;}for(var i = 0;i < intervals.Count; i++){var startInBetween = newInterval.start >= intervals[i].start && newInterval.start <= intervals[i].end;var startInGap = i < intervals.Count - 1 && newInterval.start > intervals[i].end && newInterval.start < intervals[i+1].start;//try find start fall into which interval or into which gapif(newAsStart || startInBetween || startInGap){var j = i ;// try find endvar endIndex = -1;var endInBetween = false;var endInGap = false;var endAtLast = newInterval.end > intervals[intervals.Count-1].end;if(!endAtLast){while(j < intervals.Count){if(newInterval.end >= intervals[j].start && newInterval.end <= intervals[j].end){endInBetween = true;endIndex = j;break;}else if(j < intervals.Count - 1 && newInterval.end > intervals[j].end && newInterval.end < intervals[j+1].start){endInGap = true;endIndex = j;break;}j++;}}Interval combined = null;int start = -1;int startIndex = -1;if(newAsStart){startIndex = -1;start = newInterval.start ;}else if(startInBetween){startIndex = i - 1;start = intervals[i].start;}else if(startInGap){startIndex = i;start = newInterval.start;}int end = -1;int endAt = -1;//found endif(endIndex != -1){if(endInBetween){end = intervals[endIndex].end;endAt = endIndex + 1;}else if(endInGap){end = newInterval.end;endAt = endIndex + 1;}}else if(endAtLast){end = newInterval.end;endAt = intervals.Count+1;}// not found end , means new interval end is bigger than the last end else{combined = new Interval(start, newInterval.end);endAt = intervals.Count+1;}combined = new Interval(start,end);for(var x = 0;x <= startIndex; x++){result.Add(intervals[x]);}result.Add(combined);for(var x = endAt;x < intervals.Count; x++){result.Add(intervals[(int)x]);}return result;}}//new interval start is bigger than all intervals end, just put at endresult.Add(newInterval);return result;    }}