标签：位置 container insert 长方形 white lin 范围 repeat tom
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 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 | /** * 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) { vector<Interval> result; int i; for(i = 0; i < intervals.size(); ++i){ if(intervals[i].end < newInterval.start){ result.push_back(intervals[i]); } else{ break; } } // new interval is out of the max range if(i == intervals.size()){ result.push_back(newInterval); return result; } Interval newi(min(intervals[i].start, newInterval.start), newInterval.end); for(; i < intervals.size(); ++i){ Interval curi = intervals[i]; if(curi.end < newi.end){ continue; } else{ // newi end is int the range of curi if(newi.end >= curi.start){ newi.end = curi.end; ++i; break; } //newi end is before the rnage of curi else{ break; } } } result.push_back(newi); for(;i < intervals.size(); ++i){ result.push_back(intervals[i]); } return result; }}; |
标签：位置 container insert 长方形 white lin 范围 repeat tom
原文地址：http://www.cnblogs.com/zhxshseu/p/1dcd6f5a12cb83bfed1e38c003b0bc75.html
