[hard]57. Insert Interval

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本文介绍了一个插入区间问题的解决方案，该问题要求在一个已排序且不重叠的区间集合中插入一个新的区间，并在必要时进行合并。通过具体示例展示了如何处理区间重叠的情况，并提供了一段C++代码实现。

57. 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].

Way to easy to be called hard.

class Solution {
public:
    vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
        vector<Interval> ret;
        
        bool inserted = false;

        for (auto i: intervals) {
            if (i.end < newInterval.start)
                ret.push_back(i);
            else if (i.start > newInterval.end) {
                if (!inserted)
                    ret.push_back(newInterval);
                ret.push_back(i);
                inserted = true;
            }
            else {
                if (i.end >= newInterval.start)
                    newInterval.start = min(i.start, newInterval.start);
                if (i.start <= newInterval.end)
                    newInterval.end = max(newInterval.end, i.end);
            }
        }

        if (!inserted)
            ret.push_back(newInterval);

        return ret;
    }
};