[Leetcode] 57. Insert Interval

57. 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].

给定一系列不重叠的区间，并给定一个新区间，将该区间合并插入原来的不重叠区间。

例如1：

给定区间[1,3], [6, 9], 要求插入合并区间[2, 5], 可得到新的不重叠区间[1, 5], [6, 9].

例如2 ：

给定区间[1, 2], [3, 5], [6, 7], [8, 10], [12, 16], 要求插入合并区间[4, 9], 可得到新的不重叠区间[1, 2], [3, 10], [12, 16].

解题思路

这道题的困难程度是标记为Hard的题目。一个简单的解题思路是比较要求插入合并的区间与原始不重叠区间的大小。即将最大的小于插入合并区间的区间，和最小的大于插入合并区间的区间合并起来，就可以得到新的不重叠区间。

代码实现如下：

/**
* 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) {
vector<Interval>::iterator it = intervals.begin();
while(it != intervals.end())
{
if(it->start > newInterval.end)
{
intervals.insert(it, newInterval);
return intervals;
}
else if(it->end < newInterval.start)
{
it++;
continue;
}
else
{
newInterval.start = min(newInterval.start, it->start);
newInterval.end = max(newInterval.end, it->end);
it = intervals.erase(it);
}
}
intervals.insert(intervals.end(), newInterval);
return intervals;
}
};

但是该代码算法运行时间长，无法通过LeetCode的编译，会超时。

方法二：

对以上算法进行优化，便可通过LeetCode的编译

/**
* 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) {
int sz = intervals.size();
vector<Interval> res;

for(int i = 0; i < sz; i++)
{
if(intervals[i].end < newInterval.start)
{
res.push_back(intervals[i]);
}
else if(intervals[i].start > newInterval.end)
{
res.push_back(newInterval);
res.insert(res.end(),intervals.begin()+i,intervals.end());
return res;
}
else
{
newInterval.start = min(newInterval.start, intervals[i].start);
newInterval.end = max(newInterval.end, intervals[i].end);
}
}
res.push_back(newInterval);
return res;
}
};