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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].

    先把要插入的区间放到最后,再排序,然后就跟上题一一模一样了。

     1 /**
     2  * Definition for an interval.
     3  * struct Interval {
     4  *     int start;
     5  *     int end;
     6  *     Interval() : start(0), end(0) {}
     7  *     Interval(int s, int e) : start(s), end(e) {}
     8  * };
     9  */
    10  
    11 bool cmp(const Interval &a, const Interval &b) {
    12     return a.start < b.start;
    13 }
    14 
    15 class Solution {
    16 public:
    17     vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
    18         int pos = 0, cnt = 0;
    19         intervals.push_back(newInterval);
    20         sort(intervals.begin(), intervals.end(), cmp);
    21         for (int i = 1; i < intervals.size(); ++i) {
    22             if (intervals[pos].end >= intervals[i].start) {
    23                 ++cnt;
    24                 if (intervals[pos].end < intervals[i].end) {
    25                     intervals[pos].end = intervals[i].end;
    26                 }
    27             } else {
    28                 ++pos;
    29                 intervals[pos].start = intervals[i].start;
    30                 intervals[pos].end = intervals[i].end;
    31             }
    32         }
    33         intervals.resize(intervals.size()-cnt);
    34         return intervals;
    35     }
    36 };
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  • 原文地址:https://www.cnblogs.com/easonliu/p/4218190.html
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