LeetCode题目 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].

解析：对于这个问题无非就三种情况：1、被给出的间隔的终端比要插入的起始端小，则直接将其插入。2、被给出的间隔的起始端比要插入的终端大，将两者都插入。3、介入上述两种情况之间 即存在交叉。则考虑起始和终端时应取最小的起始端和最大的终止端即可。

代码：

/** * 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) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        vector<Interval> result;    vector<Interval>::iterator it;bool flag=true;for (it=intervals.begin();it!=intervals.end();it++){if (it->end<newInterval.start){result.push_back(*it);continue;}if (it->start>newInterval.end){if (flag){result.push_back(newInterval);flag=false;}result.push_back(*it);continue;}newInterval.start=it->start<newInterval.start?it->start:newInterval.start;newInterval.end=it->end>newInterval.end?it->end:newInterval.end;}if (flag){result.push_back(newInterval);}return result;            }};