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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]
.
在原始数组上操作,先按照start值在原数组中二分查找待插入的区间,假设查找到的位置为ite,从ite或者ite-1开始合并区间直到不能合并为止(终止条件是合并后区间的end<当前区间的start),然后在原数组中删除参与合并的区间,再插入合并后的新区
间 本文地址
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class
Solution { private : static
bool comp(Interval a, Interval b) { return
a.start < b.start; } public : vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) { //在原始数组上操作 vector<Interval>::iterator ite = lower_bound(intervals.begin(),intervals.end(), newInterval, comp); //按照start值二分查找 if (ite != intervals.begin() && newInterval.start <= (ite-1)->end) //ite的上一个区间也可能参与合并 { ite--; //合并后新区间的起点只和第一个合并的区间有关,因为数组时按区间起点有序的 newInterval.start = min(newInterval.start, ite->start); } vector<Interval>::iterator eraseBegin = ite; for (; ite != intervals.end() && newInterval.end >= ite->start; ite++) if (newInterval.end < ite->end)newInterval.end = ite->end; //合并后的新区间存放于newInterval ite = intervals.erase(eraseBegin, ite); //[eraseBegin, ite)是合并时应该删掉的区间 intervals.insert(ite, newInterval); //插入合并后的区间 return
intervals; } }; |
新建数组存放结果
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/** * 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 { private : static
bool comp(Interval a, Interval b) { return
a.start < b.start; } public : vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) { vector<Interval> res; res.reserve(intervals.size()); int
i; //插入前部分不需要合并的区间 for (i = 0; i < intervals.size() && intervals[i].end < newInterval.start; i++) res.push_back(intervals[i]); //i为需要合并的起点,注意的是合并后新区间的起点只和第一个合并的区间有关,因为数组时按区间起点有序的 if (i < intervals.size())newInterval.start = min(newInterval.start, intervals[i].start); //合并区间 for (; i < intervals.size() && newInterval.end >= intervals[i].start; i++) if (newInterval.end < intervals[i].end)newInterval.end = intervals[i].end; //插入合并后的区间 res.push_back(newInterval); //插入剩余的区间 res.insert(res.end(), intervals.begin()+i, intervals.end()); return
res; } }; |
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标签:blog class code tar color int
原文地址:http://www.cnblogs.com/TenosDoIt/p/3715013.html