leetcode || 56、 Merge Intervals
2015-04-01 15:42
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problem:
Given a collection of intervals, merge all overlapping intervals.
For example,
Given
return
Hide Tags
Array Sort
题意:给定数组区间,合并有覆盖或者相邻的区间
thinking:
(1)一开始我想到用hash table的方法,开一个总区间跨度的数组,对于有区间覆盖的数组区间置为true,没被覆盖的数组区间置为false,最后将true区间的起点和终点作为区间输出即可。思路简单,但是,我忽略一个问题:区间跨度是不定的,所以要开的数组大小有可能很大。提交也显示:Memory
Limit Exceeded
(2)换一种方法,排序法。
可以直接对vector<Interval> 数组排序,要重载compare函数。也可以使用multimap<int, int>,注意不是map
code:
排序法: Accepted
Given a collection of intervals, merge all overlapping intervals.
For example,
Given
[1,3],[2,6],[8,10],[15,18],
return
[1,6],[8,10],[15,18].
Hide Tags
Array Sort
题意:给定数组区间,合并有覆盖或者相邻的区间
thinking:
(1)一开始我想到用hash table的方法,开一个总区间跨度的数组,对于有区间覆盖的数组区间置为true,没被覆盖的数组区间置为false,最后将true区间的起点和终点作为区间输出即可。思路简单,但是,我忽略一个问题:区间跨度是不定的,所以要开的数组大小有可能很大。提交也显示:Memory
Limit Exceeded
(2)换一种方法,排序法。
可以直接对vector<Interval> 数组排序,要重载compare函数。也可以使用multimap<int, int>,注意不是map
code:
排序法: Accepted
class Solution { public: vector<Interval> merge(vector<Interval> &intervals) { vector<Interval> ret; multimap<int,int> map_intervals; if(intervals.size()==0) return ret; if(intervals.size()==1) return intervals; for(vector<Interval>::iterator it=intervals.begin();it!=intervals.end();it++) map_intervals.insert(make_pair((*it).start,(*it).end)); multimap<int, int>::iterator p=map_intervals.begin(); Interval tmp(p->first,p->second); for(multimap<int, int>::iterator k=++p;k!=map_intervals.end();k++) { if(k->first<=tmp.end) tmp.end=max(tmp.end,k->second); else { ret.push_back(tmp); tmp.start=k->first; tmp.end=k->second; } } ret.push_back(tmp); return ret; } };hash table 法:Memory Limit Exceeded
class Solution { public: vector<Interval> merge(vector<Interval> &intervals) { vector<Interval> ret; int first=INT_MAX, last=INT_MIN; for(vector<Interval>::iterator tmp=intervals.begin();tmp!=intervals.end();tmp++) { first=min((*tmp).start,first); last=max((*tmp).end,last); } int count=last-first+1; bool *a = new bool[count]; memset(a,false,sizeof(bool)*count); for(vector<Interval>::iterator it=intervals.begin();it!=intervals.end();it++) { int num=(*it).end-(*it).start+1; memset(a+(*it).start,true,sizeof(bool)*num); } int interval_start=0,interval_end=0; while(interval_end<count) { interval_end=interval_start; int index=0; while(interval_start+index<count && a[index]) index++; interval_end+=index-1; Interval tmp_interval(interval_start,interval_end); ret.push_back(tmp_interval); if(interval_end<count) { int loc =interval_end+1; while(loc<count && (!a[loc])) loc++; interval_start=loc; } } } };
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