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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].
Runtime: 588ms
1 /** 2 * Definition for an interval. 3 * struct Interval { 4 * int start; 5 * int end; 6 * Interval() : start(0), end(0) {} 7 * Interval(int s, int e) : start(s), end(e) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) { 13 vector<Interval> result; 14 15 if(intervals.empty()){ 16 result.push_back(newInterval); 17 return result; 18 } 19 20 int i = 0; 21 int n = intervals.size(); 22 for(; i < n; i++){ 23 if(newInterval.end < intervals[i].start){ 24 //result.push_back(newInterval); 25 break; 26 } 27 else if(newInterval.start > intervals[i].end){ 28 result.push_back(intervals[i]); 29 continue; 30 } 31 else{ 32 newInterval.start = min(newInterval.start, intervals[i].start); 33 newInterval.end = max(newInterval.end, intervals[i].end); 34 } 35 } 36 37 result.push_back(newInterval); 38 39 for(; i < intervals.size(); i++) 40 result.push_back(intervals[i]); 41 42 return result; 43 } 44 };
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原文地址：http://www.cnblogs.com/amazingzoe/p/4851866.html
