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].

/**
* 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>::iterator itstart=intervals.begin();
for(; itstart!=intervals.end(); ++itstart)
if((*itstart).start <= newInterval.start && newInterval.start <= (*itstart).end)
break;
vector<Interval>::iterator itend=intervals.begin();
for(; itend!=intervals.end(); ++itend)
if((*itend).start <= newInterval.end && newInterval.end <= (*itend).end)
break;
if(itstart!=intervals.end() && itend!=intervals.end())
{
if(itstart != itend)
{
(*itstart).end = (*itend).end;
int tmp = (*itstart).end;
itstart++;
while(itstart != intervals.end())
if((*itstart).start <= tmp)
itstart = intervals.erase(itstart);
else break;
}
return intervals;
}else if(itstart==intervals.end() && itend!=intervals.end())
{
(*itend).start = newInterval.start;
int tmp1 = (*itend).start;
int tmp2 = (*itend).end;
itstart = intervals.begin();
while((*itstart).start != tmp1 && (*itstart).end != tmp2)
if((*itstart).start >= tmp1 && (*itstart).end <= tmp2)
itstart = intervals.erase(itstart);
else ++itstart;
return intervals;
}else if(itstart!=intervals.end() && itend==intervals.end())
{
(*itstart).end = newInterval.end;
int tmp = newInterval.end;
++itstart;
while(itstart != intervals.end())
if((*itstart).start < tmp)
itstart = intervals.erase(itstart);
else break;
return intervals;
}else{
vector<Interval>::iterator it=intervals.begin();
for(; it!=intervals.end(); ++it)
if((*it).start > newInterval.start)
break;
//--it;
vector<Interval>::iterator it1 = intervals.insert(it,newInterval);
while(it1 != intervals.end())
{
if((*it1).end < newInterval.end)
it1 = intervals.erase(it1);
else
it1++;
}
return intervals;
}
}
};

leetcode_question_57 Insert Interval的更多相关文章

  1. 【leetcode】Insert Interval

    Insert Interval Given a set of non-overlapping intervals, insert a new interval into the intervals ( ...

  2. 60. Insert Interval && Merge Intervals

    Insert Interval Given a set of non-overlapping intervals, insert a new interval into the intervals ( ...

  3. 【题解】【区间】【二分查找】【Leetcode】Insert Interval & Merge Intervals

    Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessa ...

  4. leetcode Insert Interval 区间插入

    作者:jostree  转载请注明出处 http://www.cnblogs.com/jostree/p/4051169.html 题目链接:leetcode Insert Interval 使用模拟 ...

  5. [OJ] Insert Interval

    LintCode #30. Insert Interval (Easy) LeetCode #57. Insert Interval (Hard) class Solution { public: v ...

  6. [Swift]LeetCode57. 插入区间 | Insert Interval

    Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessa ...

  7. 【LeetCode】57. Insert Interval

    Insert Interval Given a set of non-overlapping intervals, insert a new interval into the intervals ( ...

  8. Leetcode: Merge/Insert Interval

    题目 Given a collection of intervals, merge all overlapping intervals. For example,Given [1,3],[2,6],[ ...

  9. [LeetCode] Merge Interval系列,题:Insert Interval,Merge Intervals

    Interval的合并时比较常见的一类题目,网上的Amazon面经上也有面试这道题的记录.这里以LeetCode上的例题做练习. Merge Intervals Given a collection ...

随机推荐

  1. Java基础知识强化67:基本类型包装类之Integer直接赋值的面试题

    1. 面试题: Integer  i = 1: i += 1: 做了哪些事情? (1)其中Integer i =1:做了自动装箱( 使用valueOf()方法,int ---> Integer ...

  2. CentOS7--64安装python的psutil模块

    1.以root身份登陆CentOS依次 执行以下命令: wget https://pypi.python.org/packages/source/p/psutil/psutil-2.1.3.tar.g ...

  3. HTML5和CSS3实例教程[总结二]

    基于contenteditable属性实现在位编辑 HTML5规范引入了contenteditable属性,它几乎可以用在任何元素上,只要添加这一属性 即可变为可编译区域 <!DOCTYPE h ...

  4. Css的三大机制(特性):特殊性、继承、层叠详解

    继承(Inheritance)是从一个元素向其后代元素传递属性值所采用的机制.确定应当向一个元素应用那些值时,用户代理(浏览器)不仅要考虑继承,还要考虑声明的特殊性(specificity),另外需要 ...

  5. win7 重启 IIS.

    步骤 1,打开 "控制面板",并将右上角的"查看方式"设置为 "大/小图标". 2,选择 "管理工具": 3,打开 In ...

  6. 关于动态SQL中的NULL

    declare v_sql ); v_c1 number; v_c2 number; begin v_c2 :; v_sql := 'begin '; v_sql := v_sql||'update ...

  7. iOS 点击cell下拉

    iOS  点击cell下拉 代码如下: #import "ViewController.h" @interface ViewController ()<UITableView ...

  8. 不用标准库实现memmove,借助一个缓冲区temp,即使src和dest所指的内存有重叠也能正确拷贝

    void *mymemmove(void *dest,const void *src,size_t count) { char temp[count]; int i; char *d=(char *) ...

  9. (原)python中matplot中获得鼠标点击的位置及显示灰度图像

    转载请注明出处: http://www.cnblogs.com/darkknightzh/p/6182474.html 参考网址: http://matplotlib.org/examples/pyl ...

  10. mysql 5.6 参数详解

    系统变量提供的是各种与服务器配置和功能有关的信息.大部分的系统变量都可以在服务器启动时进行设置.在运行时,每一个系统变量都拥有一个全局值或会话值,或者同时拥有这两个值.许多系统变量都是动态的,也就是说 ...