57. Insert Interval

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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].

思路：

先判断是否有重叠的部分如果没有重叠的部分可以把new集合或者原来的集合i放到结果集中（看没有重叠的部分具体是什么情况）

如果有重叠的部分就更新newInterval的起止范围直到newInterval 与现存的集合没有重叠部分

/** * 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> insert(List<Interval> intervals, Interval newInterval) {       List<Interval> ret=new ArrayList<Interval>();       for(Interval i:intervals){           if(newInterval==null) {               ret.add(i);               continue;}           if(newInterval.start>i.end){               ret.add(i);               continue;           }           if(newInterval.end<i.start){               ret.add(newInterval);               ret.add(i);               newInterval=null;               continue;           }           newInterval.start=Math.min(i.start,newInterval.start);           newInterval.end=Math.max(i.end,newInterval.end);       }       if(newInterval!=null) ret.add(newInterval);       return ret;    }}

把上面的做法稍微优化了一下

/** * 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> insert(List<Interval> intervals, Interval newInterval) {       List<Interval> ret=new ArrayList<Interval>();       int i=0;       for(i=0;i<intervals.size();i++){           if(newInterval.start>intervals.get(i).end)            ret.add(intervals.get(i));            else if(newInterval.end<intervals.get(i).start)            break;            else{                newInterval.start=Math.min(newInterval.start,intervals.get(i).start);                newInterval.end=Math.max(newInterval.end,intervals.get(i).end);            }       }        ret.add(newInterval);        for(;i<intervals.size();i++)            ret.add(intervals.get(i));       return ret;    }}

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