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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].
先把要插入的区间放到最后，再排序，然后就跟上题一一模一样了。
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 11 bool cmp(const Interval &a, const Interval &b) { 12 return a.start < b.start; 13 } 14 15 class Solution { 16 public: 17 vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) { 18 int pos = 0, cnt = 0; 19 intervals.push_back(newInterval); 20 sort(intervals.begin(), intervals.end(), cmp); 21 for (int i = 1; i < intervals.size(); ++i) { 22 if (intervals[pos].end >= intervals[i].start) { 23 ++cnt; 24 if (intervals[pos].end < intervals[i].end) { 25 intervals[pos].end = intervals[i].end; 26 } 27 } else { 28 ++pos; 29 intervals[pos].start = intervals[i].start; 30 intervals[pos].end = intervals[i].end; 31 } 32 } 33 intervals.resize(intervals.size()-cnt); 34 return intervals; 35 } 36 };
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原文地址：http://www.cnblogs.com/easonliu/p/4218190.html
