leetcode - 57. Insert Interval

57. 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 wereinitially sorted according to their start times.

 

Example 1:

Given intervals [1,3],[6,9], insert andmerge [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].

 

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定义结构体interval，表示间隔。给定若干组已经排好序的间隔，然后在给一个新的间隔，最终进行merge操作后返回。

 

 

算法思想：

 

所谓的merge操作，将两个符合条件的结构体进行合并，条件为：

若结构体A的end值大于结构体B的start值；

然后A和B就要进行合并，合并成一个新的C。

C.start = min（A.start,B.start）

C.end= max（A.end,B.end）

 

遍历一次数组，然后不断地进行merge操作即可，复杂度为O（n）,因为原始数组是有序的,。

 

步骤

1.      先将新的间隔插入有序的间隔。

2.      遍历数组，进行merge操作。

/**
* 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> r;
int i;
if(intervals.size() < 1){ r.push_back(newInterval);return r;}
int endindex = intervals.size();
for(i = 0 ; i < endindex ; i++){
if(newInterval.start < intervals[i].start){
intervals.insert(intervals.begin()+i,newInterval);
break;
}//end if
}//for(i)
if(i == endindex)intervals.push_back(newInterval);
Interval now = intervals[0];
Interval temp;
for( i = 1 ; i < intervals.size() ; i++){
if(now.end < intervals[i].start ){
r.push_back(now);
now = intervals[i];
continue;
}
else{
now.end = max(now.end,intervals[i].end);
}
}//for(i)
r.push_back(now);
return r;
}
};