标签：style blog color io strong ar art div log
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].
思路：看了网上的解法，貌似直接遍历就行了。写了个二分查找的，不如直接遍历来得直观，作死了！
1 class Solution { 2 public: 3 vector<Interval> insert( vector<Interval> &intervals, Interval newInterval ) { 4 int start = binary_search( intervals, newInterval.start ); 5 if( start < 0 ) { start = 0; } 6 if( start < (int)intervals.size() && intervals[start].end < newInterval.start ) { ++start; } 7 if( start >= (int)intervals.size() ) { 8 intervals.push_back( newInterval ); 9 return intervals; 10 } 11 int end = binary_search( intervals, newInterval.end ); 12 if( end < 0 ) { 13 intervals.insert( intervals.begin(), newInterval ); 14 return intervals; 15 } 16 if( start <= end ) { 17 intervals[start].start = min( intervals[start].start, newInterval.start ); 18 intervals[start].end = max( intervals[end].end, newInterval.end ); 19 intervals.erase( intervals.begin()+start+1, intervals.begin()+end+1 ); 20 } else { 21 intervals.insert( intervals.begin()+start, newInterval ); 22 } 23 return intervals; 24 } 25 private: 26 int binary_search( vector<Interval> &intervals, int val ) { 27 int start = 0, end = (int)intervals.size()-1; 28 while( start <= end ) { 29 int middle = ( start + end ) / 2; 30 if( intervals[middle].start == val ) { return middle; } 31 if( intervals[middle].start < val ) { 32 start = middle + 1; 33 } else { 34 end = middle - 1; 35 } 36 } 37 return end; 38 } 39 };
标签：style blog color io strong ar art div log
原文地址：http://www.cnblogs.com/moderate-fish/p/3938058.html
