[LeetCode] 035: Insert Interval

[Problem]
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]

.

[Solution]

/**
* 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:
// the start of the newInterval always later than the start of the last interval in intervals
void insertInterval(vector<Interval> &intervals, Interval newInterval){
// empty
if(intervals.size() == 0){
intervals.push_back(newInterval);
}
else{
// the last of the intervals
Interval interval = intervals[intervals.size()-1];

// not intersected
if(newInterval.start > interval.end){
intervals.push_back(newInterval);
}
// intersected
else if(newInterval.start <= interval.end){
if(newInterval.end > interval.end){
intervals[intervals.size()-1].end = newInterval.end;
}
}
}
}
// insert newInterval into intervals
vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
// Start typing your C/C++ solution below
// DO NOT write int main() function

// result
vector<Interval> res;
bool inserted = false;  // whether the newInterval has been inserted
for(int i = 0; i < intervals.size(); ++i){
if(!inserted && newInterval.start <= intervals[i].start){
inserted = true;
insertInterval(res, newInterval);
}
insertInterval(res, intervals[i]);
}

// the intervals is empty or the newInterval is the last one
if(!inserted){
insertInterval(res, newInterval);
}

return res;
}
};

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