leetcode—有关区间合并

题目一：

Merge Intervals

Given a collection of intervals, merge all overlapping intervals.

For example,

Given

[1,3],[2,6],[8,10],[15,18]

,

return

[1,6],[8,10],[15,18]

.

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Array Sort

/**
* 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:
static bool cmp(const Interval& a, const Interval& b)//升序排列
{
return a.start < b.start;
}
vector<Interval> merge(vector<Interval>& intervals)
{
vector<Interval> res;
int size = intervals.size();
if (size <= 0)
return res;
sort(intervals.begin(), intervals.end(), cmp);
cout << size << endl;

res.push_back(intervals[0]);
for (int i = 0; i < size; ++i)
{
if (res.back().end >= intervals[i].start)
res.back().end = (res.back().end>intervals[i].end ? res.back().end : intervals[i].end);
else
res.push_back(intervals[i]);
}
return res;
}
};

题目二：

Insert Intervals

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]

.

class Solution {
public:
vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
vector<Interval> res;
int size = intervals.size();
if (size == 0)
{
res.push_back(newInterval);
return res;
}
int i = 0;
while (i < size)
{
if (newInterval.start>intervals[i].end)
{
res.push_back(intervals[i]);
}
else if (newInterval.end < intervals[i].start)
{
res.push_back(newInterval);
while (i < size)
{
res.push_back(intervals[i]);
i++;
}
return res;
}
else
{
newInterval.start = min(newInterval.start, intervals[i].start);
newInterval.end = max(newInterval.end, intervals[i].end);
}
i++;
}
res.push_back(newInterval);
return res;
}
};