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 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]
.
解題思路:
題意爲向一個已有的排序區間插入一個新區間,執行必要的合併,使得結構中沒有重合的區間。
解法1,先將待插入的區間插入到原來的區間數組中,然後按Merge Interval的辦法執行(http://www.kangry.net/blog/?type=article&article_id=337)。但是這樣的時間複雜度爲O(nlogn),沒有用到原區間數組已經排好序這個信息。
/**
* 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) {
intervals.push_back(newInterval);
std::sort(intervals.begin(), intervals.end(), comp);
vector<Interval> result;
int len = intervals.size();
for(int i=0; i<len; i++){
if(result.size() == 0 || !isOver(result[result.size()-1], intervals[i])){
result.push_back(intervals[i]);
}else{
result[result.size()-1].end = max(result[result.size()-1].end, intervals[i].end);
}
}
return result;
}
static bool comp(Interval& interval1, Interval& interval2){
return interval1.start < interval2.start;
}
bool isOver(Interval& interval1, Interval& interval2){
return interval1.start<=interval2.end && interval1.end>=interval2.start;
}
};
解法2、掃描區間數組,分情況討論。
(1)若新區間沒有插入,且新區間的起始節點小於當前掃描區間的起始節點,那麼將新區間插入
(2)若新區間沒有插入,且新區間的起始節點大於當前掃描區間,並且當前掃描區間與新區間沒有重合,則插入當前掃描區間
(3)若新區間沒有插入,且新區間的起始節點大於當前掃描區間,但當前掃描區間與新區間有重合,則將當前掃描區間與新區間合併後插入到結果中
(4)若新區間已經插入,則按Merge Intereval的方法處理
注意到掃描結束後還應該判斷新區間是否已經插入。
/**
* 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> result;
int len = intervals.size();
bool flag = false; //newInterval是否加入到了result中了
for(int i=0; i<len; i++){
if(!flag){
if(newInterval.start < intervals[i].start){
result.push_back(newInterval);
flag = true;
i--;
}else if(!isOver(newInterval, intervals[i])){
result.push_back(intervals[i]);
}else{
newInterval.start = min(newInterval.start, intervals[i].start);
newInterval.end = max(newInterval.end, intervals[i].end);
result.push_back(newInterval);
flag = true;
}
}else if(!isOver(result[result.size() - 1], intervals[i])){
result.push_back(intervals[i]);
}else{
result[result.size() - 1].end = max(result[result.size() - 1].end, intervals[i].end);
}
}
if(!flag){
if(result.size() == 0 || !isOver(result[result.size() - 1], newInterval)){
result.push_back(newInterval);
}else{
result[result.size() - 1].end = max(result[result.size() - 1].end, newInterval.end);
}
}
return result;
}
bool isOver(Interval& interval1, Interval& interval2){
return interval1.start<=interval2.end && interval1.end>=interval2.start;
}
};