【LeetCode】57. Insert Interval

Insert Interval

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].

由于insert和erase代价太大，需要移动后面所有元素。

所有空间换时间，返回新的数组ret，而不采用inplace做法。

主要以下三种情况：

1、newInterval与当前interval没有交集，则按照先后次序加入newInterval和当前interval，然后装入所有后续interval。返回ret。

2、newInterval与当前interval有交集，合并成为新的newInterval，然后处理后续interval。

3、处理完最后一个interval若仍未返回ret，说明newInterval为最后一个interval，装入ret。返回ret。

/**
 * 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;
        if(intervals.empty())
        {
            ret.push_back(newInterval);
            return ret;
        }

        int i = ;
        while(i < intervals.size())
        {
            //no overlapping
            if(newInterval.end < intervals[i].start)
            {
                ret.push_back(newInterval);
                while(i < intervals.size())
                {
                    ret.push_back(intervals[i]);
                    i ++;
                }
                return ret;
            }
            else if(newInterval.start > intervals[i].end)
                ret.push_back(intervals[i]);
            //overlapping
            else
            {
                newInterval.start = min(newInterval.start, intervals[i].start);
                newInterval.end = max(newInterval.end, intervals[i].end);
            }
            i ++;
        }
        ret.push_back(newInterval);
        return ret;
    }
};