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].

请看code的注释

/**
* 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:
// 判断两个区间是否interset(比较暴力)
bool checkIn(const Interval& a, const Interval& b) {
bool f1, f2;
f1 = a.start >= b.start && a.start <= b.end;
f2 = a.end >= b.start && a.end <= b.end;

bool f3, f4;
f3 = b.start >= a.start && b.start <= a.end;
f4 = b.end >= a.start && b.end <= a.end;

return f1 || f2 || f3 || f4;
}

// newInterval要么在intervals的前面, 中间, 和后面.
// 在中间的话, 要么merge, 要么独自
vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
vector<Interval> res;
int len = intervals.size();
// empty
if(len <= 0) {
res.push_back(newInterval);
return res;
}

// traverse
int i = 0;
bool f = false;
for(i = 0; i < len; i++) {
// 前面|独自|merged后的独自
if(newInterval.end < intervals[i].start) {
f = true;
break;

// merged
} else if(checkIn(intervals[i], newInterval)) {
f = true;
newInterval.start = min(newInterval.start, intervals[i].start);
newInterval.end = max(newInterval.end, intervals[i].end);
} else {
if(f) {
// has been merged
break;
}
res.push_back(intervals[i]);
}
}

// 前面|中间(独自|merged)
if(f) res.push_back(newInterval);
// the rest
for(; i < len; i++) {
res.push_back(intervals[i]);
}
// 后面
if(!f) res.push_back(newInterval);

return res;
}

};