一天一道LeetCode#57. Insert Interval
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一天一道LeetCode系列
(一)题目
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].
(二)解题
相关题目:【一天一道LeetCode】#56. Merge Intervals
/*
主要解题思路:
1、当intervals[i].end<newInterval.start的时候,不需要合并,直接i++
2、当找到第一个满足intervals[i].end>newInterval.start的时候,确定合并后的interval的start值
3、当找到最后一个满足intervals[i].start>newInterval.end的时候,确定合并后的interval的end值
*/
/**
* 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> ret;
if(intervals.size()==0){//特殊情况
ret.push_back(newInterval);
return ret;
}
int i = 0;
Interval tmp;
while(i<intervals.size()&&intervals[i].end<newInterval.start) {//将不需要合并的Interval直接压入ret
ret.push_back(intervals[i]);
i++;
}
//找到了第一个满足intervals[i].end>newInterval.start的i值
//这里需要注意当newInterval比vector里面的值都大的情况
tmp.start = min(i==intervals.size()?newInterval.start:intervals[i].start,newInterval.start);
tmp.end = newInterval.end;
while(i<intervals.size()&&intervals[i].start<=newInterval.end)
{
tmp.end = max(intervals[i].end,newInterval.end);//找到合并后的end值
i++;
}
ret.push_back(tmp);
while(i<intervals.size()) {//将剩下的Interval都压入ret
ret.push_back(intervals[i]);
i++;
}
return ret;
}
};
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