[leetcode]Merge Intervals

先写了一个O(n^2)的算法

/**
* 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> merge(vector<Interval> &intervals) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<Interval> result;
if(intervals.empty()) return result;

Interval newInterval;

for(int i = 0; i < intervals.size(); i++){
newInterval = intervals[i];
result = insert(result, newInterval);
}
return result;

}

vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<Interval> pre;
vector<Interval> post;

int low = newInterval.start;
int high = newInterval.end;

for(int i = 0; i < intervals.size(); i++){
if(intervals[i].end < low){
pre.push_back(intervals[i]);
}
if(intervals[i].start > high){
post.push_back(intervals[i]);
}
}

int p;

if(intervals.empty() || pre.size() == intervals.size()){
p = low;
}else{
p = min(low, intervals[pre.size()].start);
}

int q;
if(intervals.empty() || post.size() == intervals.size()){
q = high;
}else{
q = max(high, intervals[intervals.size()-1-post.size()].end);
}

pre.push_back(Interval(p,q));
pre.insert(pre.end(), post.begin(), post.end());

return pre;

}
};

然后翻看我以前写的，发现其实O(nlogn)就可以解决这个问题了

只要先按照start来排个序，这样每次记住前一个interval，就可以不用过去遍历之前的那些已经排好的interval了

/**
* Definition for an interval.
* struct Interval {
*     int start;
*     int end;
*     Interval() : start(0), end(0) {}
*     Interval(int s, int e) : start(s), end(e) {}
* };
*/
bool intervalLessThan(const Interval &i1, const Interval &i2){
return (i1.start < i2.start);
}

class Solution {
public:
vector<Interval> merge(vector<Interval> &intervals) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<Interval> f;
if(intervals.empty()) return f;
sort(intervals.begin(),intervals.end(),intervalLessThan);

Interval pre = intervals[0];

for(int i = 0; i < intervals.size(); i++){
Interval &cur = intervals[i];
if(pre.end < cur.start){
f.push_back(pre);
pre = cur;
continue;
}
pre.start = min(pre.start, cur.start);
pre.end = max(pre.end, cur.end);
}

if(f.empty() || f.back().end < pre.start) f.push_back(pre);

return f;

}
};

果真，第一个方法大集合100ms，第二个方法68ms