先将Intervals按start的值排列,然后从头遍历数组,将后方start小于前方end的interval相合并。如果后边元素的start大于前边元素的end,前方元素向后移一位。
时间复杂度:O(NlogN)
C++代码:
class Solution {
static bool compare(Interval i1, Interval i2)
{
return (i1.start < i2.start) ? true : false;
}
public:
vector<Interval> merge(vector<Interval>& intervals) {
if (intervals.empty())
return {};
if (intervals.size() == 1)
return intervals;
sort(intervals.begin(), intervals.end(), compare);
auto it = intervals.begin();
for (int i = 0; i < intervals.size(); i++)
{
if (it == intervals.begin() + i)
continue;
if (it->end >= intervals[i].start)
{
if (it->end < intervals[i].end)
it->end = intervals[i].end;
intervals.erase(intervals.begin() + i);
i--;
}
else
it++;
}
return intervals;
}
};