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题目：
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]. （Hard）
分析：
首先可以采用merge interval的方法，先把区间填进去，然后排序，最后再调用merge，复杂度O(NlogN)
代码：
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 private: 12 static bool cmp (const Interval& I1, const Interval& I2) { 13 return I1.start < I2.start; 14 } 15 vector<Interval> merge(vector<Interval>& intervals) { 16 vector<Interval> result; 17 if (intervals.size() == 0) { 18 return result; 19 } 20 sort(intervals.begin(), intervals.end(), cmp); 21 int left = intervals[0].start, right = intervals[0].end; 22 for (int i = 1; i < intervals.size(); ++i) { 23 if (intervals[i].start <= right) { 24 right = max(right,intervals[i].end); 25 } 26 else { 27 result.push_back(Interval(left,right)); 28 left = intervals[i].start; 29 right = intervals[i].end; 30 } 31 } 32 result.push_back(Interval(left,right)); 33 return result; 34 } 35 public: 36 vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) { 37 vector<Interval> result; 38 intervals.push_back(newInterval); 39 sort(intervals.begin(),intervals.end(),cmp); 40 result = merge(intervals); 41 return result; 42 } 43 };
当然还有O(N)的算法可以优化。
分三步来做：
第一步，找到左侧不跟newInterval相交的区间添加到结果中；
第二步，找到所有和newInterval相交的区间并找到其左边界和右边界，然后建立新的interval添加到结果中；
第三部，找到右侧不跟newInterval相交的区间添加到结果中。
注意很多细节在里面可能会犯错
代码：
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> result; 14 if (intervals.size() == 0) { 15 result.push_back(newInterval); 16 return result; 17 } 18 int i = 0; 19 while (i < intervals.size() && intervals[i].end < newInterval.start) { 20 result.push_back(intervals[i++]); 21 } 22 int left = 0; 23 if (i == intervals.size()) { 24 left = newInterval.start; 25 } 26 else { 27 left = min(newInterval.start,intervals[i].start); 28 } 29 while (i < intervals.size() && intervals[i].start <= newInterval.end) { 30 i++; 31 } 32 int right = 0; 33 if (i >= 1) { 34 right = max(newInterval.end, intervals[i - 1].end); 35 } 36 else { 37 right = newInterval.end; 38 } 39 result.push_back(Interval(left,right)); 40 while (i < intervals.size() ) { 41 result.push_back(intervals[i++]); 42 } 43 return result; 44 } 45 };
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原文地址：http://www.cnblogs.com/wangxiaobao/p/5823971.html
