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[LeetCode] Insert Interval

时间:2015-01-12 14:28:35      阅读:162      评论:0      收藏:0      [点我收藏+]

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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 };

 

[LeetCode] Insert Interval

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原文地址:http://www.cnblogs.com/easonliu/p/4218190.html

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