LeetCode 56. Merge Intervals
2016-02-10 14:04
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首先这道题的思路其实非常清晰,先根据Interval的start数值进行排序,用sort()函数,时间复杂度为O(nlgn),不需要自己去写快排什么的,直接调用sort(intervals.begin(), intervals.end(), cmp),再根据题目要求处理,注意[1,2],[1,4]归并得[1,4],而[1,6],[1,4]归并为[1,6],也就是要取end的较大值,调用max函数。
代码如下:
bool cmp(const Interval& left, const Interval& right)
{
return left.start < right.start;
}
class Solution {
public:
vector<Interval> merge(vector<Interval>& intervals)
{
if (intervals.empty())
{
return vector<Interval>();
}
sort(intervals.begin(), intervals.end(), cmp);
vector<Interval> tmp;
tmp.push_back(intervals[0]);
for (int i = 1; i < intervals.size(); ++i)
{
if (intervals[i].start > tmp.back().end)
{
tmp.push_back(intervals[i]);
}
else
{
tmp.back().end = max(intervals[i].end, tmp.back().end);
}
}
return tmp;
}
};
代码如下:
bool cmp(const Interval& left, const Interval& right)
{
return left.start < right.start;
}
class Solution {
public:
vector<Interval> merge(vector<Interval>& intervals)
{
if (intervals.empty())
{
return vector<Interval>();
}
sort(intervals.begin(), intervals.end(), cmp);
vector<Interval> tmp;
tmp.push_back(intervals[0]);
for (int i = 1; i < intervals.size(); ++i)
{
if (intervals[i].start > tmp.back().end)
{
tmp.push_back(intervals[i]);
}
else
{
tmp.back().end = max(intervals[i].end, tmp.back().end);
}
}
return tmp;
}
};
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