LeetCode（57） 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].

分析

与上一题本质相同，只需要将给定元素插入原 vector，然后将LeetCode 56题代码执行一遍即可！

AC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) {}
 * };
 */

//自定义Interval类型元素的升序比较函数
bool cmp(Interval a, Interval b)
{
    return a.start < b.start;
}

class Solution {
public:
    vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
        //如果输入参数为空，则返回空vector
        if (intervals.empty())
            return vector<Interval>(1 , newInterval);

        //将新元素插入序列中
        intervals.push_back(newInterval);

        int len = intervals.size();
        //首先，按照每个Integerval的区间首值进行排序，自定义比较
        sort(intervals.begin(), intervals.end(), cmp);

        //声明结果
        vector<Interval> ret;

        //定义临时变量
        Interval temp = intervals[0];

        for (int i = 0; i < len; i++)
        {
            //换一种判断方法
            if (intervals[i].start > temp.end)
            {
                ret.push_back(temp);
                temp = intervals[i];
            }
            else{
                temp.end = temp.end > intervals[i].end ? temp.end : intervals[i].end;
            }//else

        }//for
        ret.push_back(temp);
        return ret;
    }
};

GitHub测试程序源码