【LeetCode】Merge Intervals && Insert Interval

参考链接

Insert Interval

http://blog.csdn.net/worldwindjp/article/details/21612731
http://blog.csdn.net/littlestream9527/article/details/17686435

http://www.cnblogs.com/remlostime/archive/2012/11/25/2787387.html

题目描述

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]

.

Merge Intervals

 

Given a collection of intervals, merge all overlapping intervals.
For example,

Given 

[1,3],[2,6],[8,10],[15,18]

,

return 

[1,6],[8,10],[15,18]

.

题目分析

Merge Intervals 

 首先按start 排序 ，

 设置一个start，end

 如果 end <  intervals[i].start 就可以把 (start,end),添加进去

 如果 end <  intervals[i].end 就把end更新为  intervals[i].end，也就是把 intervals[i]合并入(start,end)

 如果 end >= intervals[i].end 说明  intervals[i] 已经在  (start,end) 不用处理 

Insert Interval

方法一，把newInterval插入按start排序的位置。然后执行merge interval
方法二，找到start和end插入的位置。见代码注释

class Solution {
public:
vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
vector<Interval> ret;
int size = intervals.size();
//TODO size 1

int i = 0;

int start = 0, end = 0;
for(i = 0;i<size;i++)
{
if(intervals[i].start> newInterval.end)//这里i后的都不用处理
break;
}
end = i;

for(;i>=0;i--)
{
if(i<size && intervals[i].end < newInterval.start)//这里i之前的都不用处理
break;
}
start = i;

for(i = 0;i<=start;i++)//start之前的肯定是和新插入的 没有交集的
{
ret.push_back(intervals[i]);
}

if(start + 1 < size && newInterval.start > intervals[start+1].start)
newInterval.start = intervals[start+1].start;

if(end>0 && newInterval.end < intervals[end-1].end)
newInterval.end = intervals[end-1].end;

ret.push_back(newInterval);

for(i = end;i<size;i++)//end之后的肯定是和新插入的 没有交集的
{
ret.push_back(intervals[i]);
}
return ret;
}
};

bool comp(const Interval &a, const Interval &b)
{
if(a.start==b.start)return a.end<b.end;
return a.start<b.start;
}
class Solution {
public:

vector<Interval> merge(vector<Interval> &intervals) {
vector<Interval> ret;
int size = intervals.size();

sort(intervals.begin(),intervals.end(),comp);

if(size < 1) return ret;
int start = intervals[0].start;
int end = intervals[0].end;
for(int i = 1;i<size;i++)
{
if(end < intervals[i].start)
{
ret.push_back(Interval(start,end));
start = intervals[i].start;
end = intervals[i].end;
}
if(end < intervals[i].end)
end = intervals[i].end;
}
ret.push_back(Interval(start,end));
return ret;
}
};

推荐学习C++的资料

C++标准函数库
http://download.csdn.net/detail/chinasnowwolf/7108919

在线C++API查询
http://www.cplusplus.com/