題目描述(Hard)

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.

題目連結

Example 1:

Input: intervals = [[1,3],[6,9]], newInterval = [2,5] Output: [[1,5],[6,9]]

Example 2:

Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8] Output: [[1,2],[3,10],[12,16]] Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].

演算法分析

基本思路就是先掃描走到新的interval應該插入的位置，接下來就是插入新的interval並檢查後面是否衝突，一直到新的interval的end小於下一個interval的start，然後取新interval和當前interval中end大的即可。因為要進行一次線性掃描，所以時間複雜度是O(n)。

提交程式碼：

/**
 * 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>::iterator iter = intervals.begin();
        while (iter != intervals.end()) {
            if (newInterval.end < iter->start) {
                intervals.insert(iter, newInterval);
                return intervals;
            } else if (newInterval.start > iter->end) {
                ++iter;
                continue;
            } else {
                newInterval.start = min(newInterval.start, iter->start);
                newInterval.end = max(newInterval.end, iter->end);
                iter = intervals.erase(iter);
            }
        }
        intervals.insert(iter, newInterval);
        return intervals;
    }
};