57. Insert Interval

Total Accepted:83528Total Submissions:314574Difficulty:HardContributors:AdminGiven a set ofnon-overlappingintervals, insert a new interval into the intervals (merge if necessary).You may assume ...

weixin_34391445

43人浏览 · 2017-02-14 21:46:00

weixin_34391445 · 2017-02-14 21:46:00 发布

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

 
          /** 
         
          * 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; 
         
          } 
         
          };