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  • LeetCode-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].

    /**
     * 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:
        bool compare(Interval& i1,Interval&i2){
            if(i1.start<i2.start)return true;
            else if(i1.start==i2.start){
                return i1.end<i2.end;
            }
            else return false;
        }
        vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
            vector<Interval> ret;
            bool put=false;
            for(int i=0;i<intervals.size();i++){
                if(intervals[i].end<newInterval.start||intervals[i].start>newInterval.end){
                    if(!put&&intervals[i].start>newInterval.end){
                        ret.push_back(newInterval);
                        put=true;
                    }
                    ret.push_back(intervals[i]);
                }
                else{
                    newInterval.start=min(newInterval.start,intervals[i].start);
                    newInterval.end=max(newInterval.end,intervals[i].end);
                }
            }
            if(!put)ret.push_back(newInterval);
            return ret;
        }
    };
    复杂度O(n)
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  • 原文地址:https://www.cnblogs.com/superzrx/p/3354552.html
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