Insert Interval
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].
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 14 int n=intervals.size(); 15 vector<Interval> result; 16 17 if(n==0) 18 { 19 result.push_back(newInterval); 20 return result; 21 } 22 23 int start=newInterval.start; 24 int end=newInterval.end; 25 26 27 bool isfindStart=false; 28 bool isfindEnd=false; 29 bool addCur=true; 30 31 for(int i=0;i<n;i++) 32 { 33 addCur=true; 34 35 if(!isfindStart) 36 { 37 //和当前区间没有交叉,且在start的前面 38 if(start>intervals[i].end) 39 { 40 //需要添加当前元素 41 addCur=true; 42 //当前元素为最后一个元素时，则认为start找到了 43 if(i==n-1) isfindStart=true; 44 } 45 else 46 { 47 //和当前区间有交叉 或者start小于当前区间 48 49 start=min(start,intervals[i].start); 50 //后续就不用再找start了 51 isfindStart=true; 52 53 //如果start和end没有和其他Interval有交叉 54 if(end<intervals[i].start) addCur=true; 55 else addCur=false; 56 } 57 } 58 59 60 if(!isfindEnd) 61 { 62 //需要继续找end 63 if(end>intervals[i].end) 64 { 65 66 if(start<=intervals[i].end) addCur=false; 67 68 if(i==n-1) 69 { 70 isfindEnd=true; 71 72 if(addCur) result.push_back(intervals[i]); 73 result.push_back(Interval(start,end)); 74 75 continue; 76 } 77 } 78 else 79 { 80 //end小于当前区间，此时找到了end 81 end=end>=intervals[i].start?intervals[i].end:end; 82 83 //找到了end 84 isfindEnd=true; 85 if(end>=intervals[i].start) addCur=false; 86 87 result.push_back(Interval(start,end)); 88 if(addCur) result.push_back(intervals[i]); 89 90 continue; 91 } 92 } 93 94 if(addCur) result.push_back(intervals[i]); 95 } 96 return result; 97 } 98 };