标签:style color io os ar for sp on art
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]
.
/** * Definition for an interval. * struct Interval { * int start; * int end; * Interval() : start(0), end(0) {} * Interval(int s, int e) : start(s), end(e) {} * }; */ struct Interval { int start; int end; Interval() : start(0), end(0) {} Interval(int s, int e) : start(s), end(e) {} }; bool operator <(const Interval a, const Interval b) { return a.start < b.start; } class Solution { public: std::vector<Interval> insert(std::vector<Interval> &intervals, Interval newInterval) { std::vector<Interval> res; intervals.push_back(newInterval); std::sort(intervals.begin(),intervals.end()); int left = intervals[0].start, right = intervals[0].end; for (int i = 1; i < intervals.size(); i++) { if(intervals[i].start <= right) { right = std::max(right,intervals[i].end); } else { res.push_back(Interval(left,right)); left = intervals[i].start; right = intervals[i].end; } } res.push_back(Interval(left,right)); return res; } };
标签:style color io os ar for sp on art
原文地址:http://blog.csdn.net/akibatakuya/article/details/40476987