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].
给定一个不重叠区间集合，集合按区间的起始值已经排好序，插入一个区间，如果插入的区间与某些区间重叠，则合并
依次扫描集合中的各个区间，和待插入集合比较，确定待插入集合的位置或新合并集合的起始值。
/**
* Definition for an interval.
* struct Interval {
* int start;
* int end;
* Interval() : start(0), end(0) {}
* Interval(int s, int e) : start(s), end(e) {}
* };
*/
class Solution {
public:
vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
vector<Interval>result;
int size=intervals.size();
bool hasInserted=false; //是否已经插入
for(int i=0; i<size; i++){
if(hasInserted){
//如果新区间已经插入，则区间集合中的后续集合直接添加到result集合中即可
result.push_back(intervals[i]);
continue;
}
if(intervals[i].start<newInterval.start){
if(intervals[i].end<newInterval.start){
//在新区间的左侧
result.push_back(intervals[i]);
}
else if(intervals[i].end<=newInterval.end){
//右边界新区间内，也即存在重合，更新新区间的左边界
newInterval.start=intervals[i].start;
}
else{
//当前区间覆盖了新区间
result.push_back(intervals[i]);
hasInserted=true;
}
}
else{
if(intervals[i].start>newInterval.end){
//在新区间的右边，插入新区间，及当前区间 【注意，这种情况下需要插入两个区间】
result.push_back(newInterval);
result.push_back(intervals[i]);
hasInserted=true;
}
else if(intervals[i].end<=newInterval.end){
//当前区间被新区间包含
continue;
}
else{
//左边界在新区间内，也即存在重合，更新新区间的右边界
newInterval.end=intervals[i].end;
}
}
}
if(!hasInserted){
result.push_back(newInterval);
}
return result;
}
};LeetCode: Insert Interval [056],布布扣,bubuko.com
LeetCode: Insert Interval [056]
原文地址：http://blog.csdn.net/harryhuang1990/article/details/26809453
