LeetCode #57 - Insert Interval

最新推荐文章于 2020-01-22 07:27:16 发布

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本文深入解析插入区间算法，介绍如何在已排序的不重叠区间集合中插入新区间并进行必要合并。通过实例演示，解释算法流程及实现细节，帮助读者理解区间合并的逻辑。

题目描述：

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:

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].

由于区间都已经排好序，所以从第一个区间开始，判断是否和插入区间相交，直到找到相交的区间之后进行合并，将合并的新区间和下一个区间比较，直到不再相交为止。

/**
 * 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> result;
        int i=0;
        while(i<intervals.size())
        {
            if(intervals[i].end<newInterval.start) 
            {
                result.push_back(intervals[i]);
                i++;
            }
            else break;
        }
        while(i<intervals.size()&&intersect(intervals[i],newInterval))
        {
            newInterval.start=min(newInterval.start,intervals[i].start);
            newInterval.end=max(newInterval.end,intervals[i].end);
            i++;
        }
        result.push_back(newInterval);
        while(i<intervals.size())
        {
            result.push_back(intervals[i]);
            i++;
        }
        return result;
    }
    
    bool intersect(Interval a, Interval b)
    {
        if(a.end<b.start||b.end<a.start) return false;
        else return true;
    }
};