算法第六週作業01

Description

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

• 順序遍歷輸入intervals中的所有元素interval，將interval與輸入的newInterval比較
• 根據interval和newInterval的左右邊界分情況討論

Code

public class Solution {
    public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
        List<Interval> result = new ArrayList<Interval>(intervals.size());
        Interval interval, tmp = null;
        for (int i = 0; i < intervals.size(); i++) {
            interval = intervals.get(i);
            if (interval.end < newInterval.start) {
                result.add(interval);
            } else if (interval.start <= newInterval.end) {
                if (tmp == null) {
                    tmp = new Interval(interval.start < newInterval.start ? interval.start : newInterval.start,
                            interval.end > newInterval.end ? interval.end : newInterval.end);
                } else {
                    tmp.end = interval.end > newInterval.end ? interval.end : newInterval.end;
                }
            } else {
                if (tmp != null) {
                    result.add(tmp);
                } else {
                    result.add(newInterval);
                }
                result.addAll(intervals.subList(i, intervals.size()));
                return result;
            }
        }
        if (tmp != null)
            result.add(tmp);
        else 
            result.add(newInterval);
        return result;
    }
}