Insert Interval

一. Insert Interval

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].

Difficulty:Hard

TIME：TIMEOUT

解法

这道题的题意很简单，但是也不是那么容易就能写出正确而且简洁的代码。

其实这道题大约可以分为三种情况，第一种就是overlap之前，第二种就是overlap之中，第三种就是overlap之后。

一开始我想的是通过一个for循环来判断各种情况，不过后面发现那样写太过复杂。因此改成了三个循环的模式，分别对应三种情况，这样判断的条件就会少很多。

/** * Definition for an interval. * struct Interval { *     int start; *     int end; *     Interval() : start(0), end(0) {} *     Interval(int s, int e) : start(s), end(e) {} * }; */vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {    vector<Interval> result;    int i = 0;    /*第一次循环，跳过overlap之前的数据*/    while(i < intervals.size() && newInterval.start > intervals[i].end) {        result.push_back(intervals[i]);        i++;    }    result.push_back(newInterval);    /*第二次循环，overlap的处理*/    while(i < intervals.size() && newInterval.end >= intervals[i].start) {        result.back().start = min(intervals[i].start, result.back().start);        result.back().end = max(intervals[i].end, result.back().end);        i++;    }    /*第三次循环，跳过overlap之后的数据*/    while(i < intervals.size()) {        result.push_back(intervals[i]);        i++;    }    return result;}

代码的时间复杂度为O(n)。