Given a collection of intervals, merge all overlapping intervals.
Example 1:
Input: [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].
Example 2:
Input: [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.
题解:
把数据排序之后遍历就行了
class Solution {
public:
static bool cmp (Interval a, Interval b) {
if (a.start != b.start) {
return a.start < b.start;
}
else {
return a.end < b.end;
}
}
vector<Interval> merge(vector<Interval>& intervals) {
int n = intervals.size();
if (n == 0) {
return intervals;
}
sort(intervals.begin(), intervals.end(), cmp);
vector<Interval> ans;
ans.push_back(intervals[0]);
for (int i = 1; i < n; i++) {
if (intervals[i].start <= ans.back().end) {
ans.back().end = max(ans.back().end, intervals[i].end);
}
else {
ans.push_back(intervals[i]);
}
}
return ans;
}
};