Merge Intervals 题解

题目来源:Merge Intervals

> Given a collection of intervals, merge all overlapping intervals. For example, Given [1,3],[2,6],[8,10],[15,18], return [1,6],[8,10],[15,18].

解题思路:

先对start排序, 排序时这样 [1,4] [1,5]时,[1,5][1,4] 统一处理[1,4][2,4]吃掉。 然后数轴上对前后两个interval作讨论,3种情况,一种including,i+1包含在i里面; i+1和i相交,i+1和i相隔。

    /**
     * Definition for an interval.
     * struct Interval {
     *     int start;
     *     int end;
     *     Interval() : start(0), end(0) {}
     *     Interval(int s, int e) : start(s), end(e) {}
     * };
     */
    int cmp(const Interval &i1, const Interval &i2)
    {
        if(i1.start == i2.start) return i1.end > i2.end; // make second bigger first
        return i1.start < i2.start;
    }
    class Solution 
    {
    public:
        vector<Interval> merge(vector<Interval> &intervals)
        {
            vector<Interval> result;
            if(intervals.size() <= 1) return intervals;
            std::sort(intervals.begin(), intervals.end(), cmp);
            int i = 1;
            auto start = intervals[0];
            while(i < intervals.size())
            {
                if(intervals[i].end <= start.end)
                {
                    i++;
                }else if(intervals[i].start <= start.end && intervals[i].end >= start.end)
                {
                    start.end = intervals[i].end;
                    i++;
                }
                else
                {
                    result.push_back(start);
                    if(i < intervals.size())
                        start = intervals[i];
                    else
                        return result;
                }
            }
            result.push_back(start);
            return move(result);
        }
    };

当然,也可以借助 Insert Interval 一个一个插入到结果里面去。

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