Insert Interval
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.
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].
分析
无
代码
- Java
- C++
// Insert Interval
// 时间复杂度O(n)，空间复杂度O(1)
public class Solution {
public int[][] insert(int[][] intervals, int[] newInterval) {
ArrayList<int[]> list = new ArrayList<>(Arrays.asList(intervals));
insert(list, newInterval);
return list.toArray(new int[0][2]);
}
private void insert(ArrayList<int[]> intervals, int[] newInterval) {
for (int i = 0; i < intervals.size();) {
final int[] cur = intervals.get(i);
if (newInterval[1] < cur[0]) {
intervals.add(i, newInterval);
return;
} else if (newInterval[0] > cur[1]) {
++i;
} else {
newInterval[0] = Math.min(newInterval[0], cur[0]);
newInterval[1] = Math.max(newInterval[1], cur[1]);
intervals.remove(i);
}
}
intervals.add(newInterval);
}
}
// Insert Interval
// 时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
vector<Interval>::iterator it = intervals.begin();
while (it != intervals.end()) {
if (newInterval.end < it->start) {
intervals.insert(it, newInterval);
return intervals;
} else if (newInterval.start > it->end) {
it++;
continue;
} else {
newInterval.start = min(newInterval.start, it->start);
newInterval.end = max(newInterval.end, it->end);
it = intervals.erase(it);
}
}
intervals.insert(intervals.end(), newInterval);
return intervals;
}
};