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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].
/** * 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 { public: vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) { int intervalsSize = intervals.size(); vector<Interval> res; if(intervalsSize==0){ res.push_back(newInterval); return res; } bool isFindStart = false,isFindEnd = false; int start = newInterval.start,end = newInterval.end; int startIndex = 0,endIndex=0; for(int i =0;i<intervalsSize;i++){//查找合并区间的起始和结束位置 if(!isFindStart){ if(start >= intervals[i].start && start<=intervals[i].end){ isFindStart = true; start = intervals[i].start; startIndex = i; }else if(start<intervals[i].start){ startIndex = i; isFindStart = true; } } // cout<<"end="<<end<<",intervals["<<i<<"].start="<<intervals[i].start<<",intervals["<<i<<"].end="<<intervals[i].end<<endl; if(!isFindEnd){ if(end >= intervals[i].start && end<=intervals[i].end ){ isFindEnd = true; end = intervals[i].end; endIndex = i; }else if(end<intervals[i].start){ endIndex = i-1; isFindEnd = true; } } if(isFindStart && isFindEnd){ break; } } if(!isFindStart) startIndex = intervalsSize; for(int j=0;j<startIndex;j++){ res.push_back(intervals[j]); } res.push_back(Interval(start,end)); if(!isFindEnd) endIndex = intervalsSize-1; for(int j=endIndex+1;j<intervalsSize;j++){ res.push_back(intervals[j]); } return res; } };
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原文地址：http://www.cnblogs.com/zengzy/p/5006005.html
