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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]
.
醉了,代码写的太乱,不过还是ac了
#include<iostream> #include<vector> #include<algorithm> #include<functional> using namespace std; struct Interval { int start; int end; Interval() : start(0), end(0) {} Interval(int s, int e) : start(s), end(e) {} }; struct overlap: public binary_function<Interval,Interval,bool> { bool operator()(const Interval &x, const Interval &y)const{ if (y.start > x.end || y.end < x.start) return false; else return true; } }; struct myless : public binary_function<Interval, Interval, bool> { bool operator()(const Interval &x, const Interval &y)const{ if (y.start > x.start) return false; else return true; } }; vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) { vector<Interval> Result; if (intervals.empty()){ Result.push_back(newInterval); return Result; } //寻找不重合的第一个区间. vector<Interval>::iterator iter_begin = find_if(intervals.begin(), intervals.end(), bind2nd(overlap(), newInterval)); Result.assign(intervals.begin(), iter_begin); if (iter_begin==intervals.end()) { auto it = find_if(Result.begin(),Result.end(), bind2nd(myless(), newInterval)); Result.insert(it, newInterval); return Result; } //确定重合区间的头尾 Interval *merge = new Interval(); if (newInterval.start < (*iter_begin).start) merge->start = newInterval.start; else merge->start = (*iter_begin).start; int end_tmp = 0; vector<Interval>::iterator iter = iter_begin; for (;iter != intervals.end(); ++iter){ if (iter->start>newInterval.end){ merge->end = end_tmp; break; } else end_tmp = max(newInterval.end, iter->end); } if (iter == intervals.end()) merge->end = end_tmp; Result.push_back(*merge); //插入余下的区间 Result.insert(Result.end(), iter, intervals.end()); return Result; }
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原文地址:http://blog.csdn.net/li_chihang/article/details/44180849