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  • Insert Interval

    在已经排好序的区间中,插入一个新的区间,与merge的做法类似

    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:
        vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
            vector<Interval> res;
            int size = intervals.size();
            int i = 0;
            for (i = 0; i < size; i++) {
                if(newInterval.end < intervals[i].start){
                    break;
                }else if(newInterval.start > intervals[i].end){
                    res.push_back(intervals[i]);
                    continue;
                }else{
                    newInterval.start = min(newInterval.start,intervals[i].start);
                    newInterval.end = max(newInterval.end,intervals[i].end);
                }
            }
            res.push_back(newInterval);
            for(int j = i;j<size;j++){
                res.push_back(intervals[j]);
            }
            return res;
        }
    };
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  • 原文地址:https://www.cnblogs.com/wxquare/p/6054266.html
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