[Leetcode] 57. Insert Interval 解题报告

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].

思路：

这道题目和Leetcode 56的思路比较类似，主要算法分为三个步骤：1）在结果集中添加好所有和newInterval没有重合，并且处在newInterval前面的区间；2）将newInterval添加到结果集，并且不断合并和newInterval有overlap的区间；3）在结果集中添加所有和newInterval没有重合，并且处在newInterval后面的区间。整个算法的时间复杂度也为O(n)，空间复杂度为O(1)。

代码：

/** * Definition for an interval. * 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;        int start = 0;        while(start < intervals.size() && newInterval.start > intervals[start].end)     // add the previous ones            ret.push_back(intervals[start++]);        while(start < intervals.size() && newInterval.end >= intervals[start].start)    // merge the overlapped ones        {            newInterval.start = min(newInterval.start, intervals[start].start);            newInterval.end = max(newInterval.end, intervals[start++].end);        }        ret.push_back(newInterval);        while(start < intervals.size())                                                 // add the subsequent ones            ret.push_back(intervals[start++]);        return ret;    }};