标签：style blog color io ar for strong sp div
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].
向已经有序的且没重合的区间集合中插入新区间，并合并由此引来的重合区间，输出。
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> ans; //存放结果 14 int i=0; 15 while( i<intervals.size() && intervals[i].end < newInterval.start ) //如果旧区间的end小于新区间的start，那么直接放入结果中 16 ans.push_back( intervals[i++] ); 17 if( i == intervals.size() ) { //处理特定的情况，及新区间的start比旧区间最后一个区间的end还大 18 ans.push_back(newInterval); 19 return ans; 20 } 21 //不断合并重复区间，直到第一个旧区间的start大于新区间的end 22 while( i<intervals.size() && intervals[i].start <= newInterval.end ) { 23 newInterval.start = min(intervals[i].start, newInterval.start); 24 newInterval.end = max(intervals[i].end, newInterval.end); 25 ++i; 26 } 27 ans.push_back(newInterval); 28 while( i<intervals.size() ) ans.push_back(intervals[i++]); //将后面不需要合并的区间放入结果中 29 return ans; 30 } 31 };
标签：style blog color io ar for strong sp div
原文地址：http://www.cnblogs.com/bugfly/p/4008516.html
