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].

Solution: Using sorting

static bool myComp(Interval a, Interval b) {
    return a.start < b.start;
}

vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
    intervals.push_back(newInterval);
    sort(intervals.begin(), intervals.end(), myComp);

    vector<Interval> ans;
    for (auto interval: intervals) {
        if (ans.empty() || ans.back().end < interval.start) {
            ans.push_back(interval);
        } else {
            ans.back().end = max(ans.back().end, interval.end);
        }
    }
    return ans;
}

Solution:

具体思路是，我们用一个变量cur来遍历区间，如果当前cur区间的结束位置小于要插入的区间的起始位置的话，说明没有重叠，则将cur区间加入结果res中，然后cur自增1。直到有cur越界或有重叠while循环退出，然后再用一个while循环处理所有重叠的区间，每次用取两个区间起始位置的较小值，和结束位置的较大值来更新要插入的区间，然后cur自增1。直到cur越界或者没有重叠时while循环退出。之后将更新好的新区间加入结果res，然后将cur之后的区间再加入结果res中即可，参见代码如下：

vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
    vector<Interval> res;
    int n = intervals.size(), cur = 0;

    while (cur < n && intervals[cur].end < newInterval.start) {
        res.push_back(intervals[cur++]);
    }
    while (cur < n && intervals[cur].start <= newInterval.end) {
        newInterval.start = min(newInterval.start, intervals[cur].start);
        newInterval.end = max(newInterval.end, intervals[cur].end);
        ++cur;
    }

    res.push_back(newInterval);
    while (cur < n) {
        res.push_back(intervals[cur++]);
    }
    return res;
}