Insert -interval
Title description
Given a set of non-overlapping time intervals, insert a new time interval in the time interval (if there is overlap, merge the intervals).
These time intervals are initially sorted according to their start time.
Example 1:
Given a time interval [1,3], [6,9], insert a time interval [2,5] in these two time intervals, and merge it with the original time interval to become [1 ,5],[6,9].
Example 2:
Given time interval [1,2],[3,5],[6,7],[8,10],[12,16], at these times Insert the time interval [4,9] into the interval and merge it with the original time interval to become [1,2],[3,10],[12,16].
This is because the time interval [4, 9] covers the time interval [3,5],[6,7],[8,10].
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].
Problem-solving ideas
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Drawing lessons from the idea of merging intervals , since the intervals are initially ordered, the intervals array is traversed to find the correct insertion position of newInterval. After inserting, it becomes a problem of merging interval, just apply the template directly.
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Usage of insert() in vector:
insert(it, x) is used to insert an element x to any iterator it of the vector, and the time complexity is O(N).
vector<int> vi; for (int i = 1; i <= 5; i++) { vi.push_back(i); //此时为1 2 3 4 5 } vi.insert(vi.begin() + i, -1); //将－1插入vi中数组下标为2的位置 /* 输出结果为1 2 -1 3 4 5 */Therefore, inserting the element x into the array at the position where the index is i is written as:
it.insert(it.begin() + i, -1) -
The code for this question is as follows:
/**
* 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> res;
if(!intervals.size()) {
res.push_back(newInterval);
return res;
}
//由于intervals初始有序，因此遍历数组，寻找newInterval正确的插入位置
for(int i = 0; i < intervals.size(); i ++) {
if(newInterval.start <= intervals[i].start) {
intervals.insert(intervals.begin() + i, newInterval);
break;
}
}
//如果newInterval比所有区间都大，则直接添加在最后
if(newInterval.start > intervals.back().start) {
intervals.push_back(newInterval);
}
//以下为区间合并的模板
res.push_back(intervals[0]);
for(int i = 1; i < intervals.size(); i ++) {
Interval temp = intervals[i];
//如果存在重叠
if(res.back().end >= temp.start) {
res.back().end = max(res.back().end, temp.end);
} else {
res.push_back(temp);
}
}
return res;
}
};