1. 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].
2. 思路
序列自身的递增且无交的。
找到待插入位置,然后对这个位置开始往后进行merge。
3. 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) {}
* };
*/
class Solution {
public:
vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
auto it = intervals.begin();
while (it != intervals.end()) {
if (newInterval.start > (*it).end) {
it++;
} else {
break;
}
}
if (it == intervals.end()) {
intervals.push_back(newInterval);
return intervals;
}
// now: it.end >= newInterval.start
if ((*it).start > newInterval.start) {
it = intervals.insert(it, newInterval);
} else {
(*it).end = max((*it).end, newInterval.end);
}
merge(intervals, it);
return intervals;
}
void merge(vector<Interval>& ivs, vector<Interval>::iterator beg) {
auto it = beg;
while (it != ivs.end()) {
if (beg->end >= it->end) {
it++;
continue;
} else if (beg->end >= it->start) {
beg->end = it->end;
it++;
}
break;
}
ivs.erase(beg+1, it);
return ;
}
};
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