来源：Leetcode
原帖：
http://oj.leetcode.com/problems/insert-interval/
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].
Solution: For example 2:
1. compare [1,2] with [4,9], then insert [1,2];
2. merge [3,5] with [4,9], get newInterval = [3,9];
3. merge [6,7] with [3,9], get newInterval = [3,9];
4. merge [8,10] with [3,9], get newInterval = [3,10];
5. compare [12,16] with [3,10], insert newInterval [3,10], then all the remaining intervals...
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) {
vector<Interval> res;
if (intervals.empty()) {
res.push_back(newInterval);
return res;
}
bool inserted = false;
for (int i = 0; i < intervals.size(); ++i) {
if (inserted || intervals[i].end < newInterval.start) { // non-overlaping
res.push_back(intervals[i]);
} else if (newInterval.end < intervals[i].start) {
res.push_back(newInterval);
res.push_back(intervals[i]);
inserted = true;
} else { // update new interval
newInterval.start = min(intervals[i].start, newInterval.start);
newInterval.end = max(intervals[i].end, newInterval.end);
}
}
if (!inserted) res.push_back(newInterval);
return res;
}
};
