【Leetcode】【Hard】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].

解题思路：

1、先用两次二分查找找到和待插入的区间有关系的区间范围；

2、将有关系的区间做处理；

3、生成新的合并后的区间并返回；

解题步骤：

代码：

 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 class Solution {
11 public:
12     static bool partial_order(const Interval &a, const Interval &b) {
13         return a.end < b.start;
14     };
15
16     vector<Interval> insert(std::vector<Interval> &intervals, Interval newInterval) {
17         vector<Interval>::iterator less = lower_bound(intervals.begin(), intervals.end(), newInterval, partial_order);
18         vector<Interval>::iterator greater = upper_bound(intervals.begin(), intervals.end(), newInterval, partial_order);
19
20         vector<Interval> answer;
21         answer.insert(answer.end(), intervals.begin(), less);
22         if (less < greater) {
23             newInterval.start = min(newInterval.start, (*less).start);
24             newInterval.end = max(newInterval.end, (*(greater - 1)).end);
25         }
26         answer.push_back(newInterval);
27         answer.insert(answer.end(), greater, intervals.end());
28
29         return answer;
30   }
31 };

不用lower_bound和upper_bound写的AC烂代码，留作改进：

 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 class Solution {
11 public:
12     vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
13         if (intervals.empty())
14             return vector<Interval> (1, newInterval);
15         int left = 0;
16         int right = intervals.size() - 1;
17         int size = intervals.size();
18         int ins_start, ins_end;
19
20         while (left < right) {
21             int mid = (left + right) / 2;
22             if (intervals[mid].end < newInterval.start)
23                 left = mid + 1;
24             else
25                 right = mid;
26         }
27         ins_start = left;
28
29         left = left == 0 ? 0 : left - 1;
30         right = intervals.size() - 1;
31         while (left < right) {
32             int mid = (left + right) / 2 + 1;
33             if (newInterval.end < intervals[mid].start)
34                 right = mid - 1;
35             else
36                 left = mid;
37         }
38         ins_end = right;
39
40         if (ins_end == 0 && newInterval.end < intervals[0].start) {
41             intervals.insert(intervals.begin(), newInterval);
42             return intervals;
43         }
44         if (ins_start == size - 1 && newInterval.start > intervals[size - 1].end) {
45             intervals.push_back(newInterval);
46             return intervals;
47         }
48
49         vector<Interval> answer;
50         answer.insert(answer.end(), intervals.begin(), intervals.begin() + ins_start);
51         if (ins_start <= ins_end) {
52             newInterval.start = min(newInterval.start, intervals[ins_start].start);
53             newInterval.end = max(newInterval.end, intervals[ins_end].end);
54         }
55         answer.push_back(newInterval);
56         answer.insert(answer.end(), intervals.begin() + ins_end + 1, intervals.end());
57         return answer;
58
59     }
60 };