【LeetCode】Insert Interval

题目描述：

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].

如果在原数组里插入的话，会有重新排序的问题，每一次insert操作都是O（n）的复杂度，索性不如直接返回新数组。

代码：

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> ret;if (intervals.empty())return vector<Interval>(1,newInterval);int i(0);bool flag(false);while (i < intervals.size()){Interval curr = intervals[i];if (!flag&&newInterval.start < intervals[i].start){curr = newInterval;flag = true;}elsei++;if (ret.empty()){ret.push_back(curr);continue;}Interval *prev = &ret[ret.size() - 1];if (curr.start>prev->end)ret.push_back(curr);elseprev->end = max(prev->end, curr.end);}if (!flag){Interval *prev = &ret[ret.size() - 1];if (newInterval.start>prev->end)ret.push_back(newInterval);elseprev->end = max(prev->end, newInterval.end);}return ret;}};