Merge Intervals 题解
题目来源: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].
解题思路:
先对start排序, 排序时这样 [1,4] [1,5]时,[1,5][1,4] 统一处理[1,4][2,4]吃掉。 然后数轴上对前后两个interval作讨论,3种情况,一种including,i+1包含在i里面; i+1和i相交,i+1和i相隔。
/**
* Definition for an interval.
* struct Interval {
* int start;
* int end;
* Interval() : start(0), end(0) {}
* Interval(int s, int e) : start(s), end(e) {}
* };
*/
int cmp(const Interval &i1, const Interval &i2)
{
if(i1.start == i2.start) return i1.end > i2.end; // make second bigger first
return i1.start < i2.start;
}
class Solution
{
public:
vector<Interval> merge(vector<Interval> &intervals)
{
vector<Interval> result;
if(intervals.size() <= 1) return intervals;
std::sort(intervals.begin(), intervals.end(), cmp);
int i = 1;
auto start = intervals[0];
while(i < intervals.size())
{
if(intervals[i].end <= start.end)
{
i++;
}else if(intervals[i].start <= start.end && intervals[i].end >= start.end)
{
start.end = intervals[i].end;
i++;
}
else
{
result.push_back(start);
if(i < intervals.size())
start = intervals[i];
else
return result;
}
}
result.push_back(start);
return move(result);
}
};当然,也可以借助 Insert Interval 一个一个插入到结果里面去。
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