57. Insert Interval

• Total Accepted: 83528
• Total Submissions: 314574
• Difficulty: Hard
• Contributors: Admin

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].

来源： https://leetcode.com/problems/insert-interval/

分析

---

把插入newInterval看成如下形式，想象成把一个长方形的容器倒盖在现有的排好顺序的容器上

  |~~~~~~~~~~~~~~~~~~~~~~|       newInterval

|____| |__| |________|  |______| intervals

先确定newInterval的 start 在哪个位置，假设当前指向的intervals[i] 叫做 curi

如果newInterval.end > curi.end

那么将该curi 压入结果 result，并继续对下一个interval进行判断(可能会出现newInterval超出边界的情况，则直接压入结果result的最末端)

否则：

1 在curi的内部，即满足 newInterval.start >= curi.start 

     |~~~~~~~~~~~~~~......... newInterval 

...|____| |__| |_______...... intervals

2 在curi的前面的空隙，即满足 newInterval.start < curi.start

      |~~~~~~~~~~~~~~~~~~......... newInterval 

..._|   |____| |__| |_______...... intervals

确定新interval，即newi的start，只需要取 curi.start 和 newInterval.start的最小值就行

先给newi的end赋值 newInterval.end，然后不断跳过那些在 newInterval.end 范围内的 intervals[i], 直到curi.end >= newInterval.end，如下图所示两种情况：

.....~~~~~~~~~~~~~~~~~~~~|       newInterval

..|____| |__| |________|  |______| intervals

                            curi

.....~~~~~~~~~~~~~~~~~~~~~~~~~|    newInterval

..|____| |__| |________|  |______| intervals

                            curi

分别对这两种情况确定 newi.end

最后将剩余的intervals[i] 压入结果result

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54

/**  * 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> result;         int i;         for(i = 0; i < intervals.size(); ++i){             if(intervals[i].end < newInterval.start){                 result.push_back(intervals[i]);             }             else{                 break;             }         }         // new interval is out of the max range         if(i == intervals.size()){             result.push_back(newInterval);             return result;         }                   Interval newi(min(intervals[i].start, newInterval.start), newInterval.end);         for(; i < intervals.size(); ++i){             Interval curi = intervals[i];             if(curi.end < newi.end){                 continue;             }             else{                 // newi end is int the range of curi                 if(newi.end >= curi.start){                     newi.end = curi.end;                     ++i; break;                 }                 //newi end is before the rnage of curi                 else{                     break;                 }             }         }         result.push_back(newi);                   for(;i < intervals.size(); ++i){             result.push_back(intervals[i]);         }         return result;     } };

null