LeetCode 57. Insert Interval（插入区间）

题目描述：    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.

例一：Given intervals 

[1,3],[6,9]

, insert and merge 

[2,5]

 in as 

[1,5],[6,9]

.
例二：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]

.
分析：
    题意：给出一些非重叠整型区间集合，插入一个新的区间，返回合并重叠区间之后的结果。
    思路：这道题跟LeetCode 56很相似，我们用一个小技巧进行转化：先把新的区间插入到所有的区间集合中去，根据区间end进行从小到大的排序，然后采用LeetCode 56的贪心算法来解决这道题。因为思路完全一致，这里不再复述算法细节。

代码：
#include <bits/stdc++.h>

using namespace std;

struct Interval{
int start;
int end;
Interval(): start(0), end(0){}
Interval(int s, int e): start(s), end(e){}
};

class Solution {
private:
static int cmp(const Interval a, const Interval b){
if(a.end != b.end){
return a.end < b.end;
}
return a.start < b.start;
}

public:
vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
vector<Interval> ans;
intervals.push_back(newInterval);
int n = intervals.size();
// sort
sort(intervals.begin(), intervals.end(), cmp);
for(int i = 1; i <= n - 1; i++){
int j = i - 1;
while(j >= 0){
if(intervals[j].start == -1 && intervals[j].end == -1){
j--;
continue;
}
if(intervals[i].start <= intervals[j].end){
intervals[i].start = min(intervals[j].start, intervals[i].start);
intervals[j].start = intervals[j].end = -1;
j--;
}
else{
break;
}
}
}
// get answer
for(Interval p: intervals){
if(p.start == -1 && p.end == -1){
continue;
}
ans.push_back(p);
}
return ans;
}
};