LeetCode | Insert Interval

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].

从头到尾来一遍，O(n)的复杂度。

看到网上还有其他比较凶残的O(lgn)的方法，没有细看了。Java版本用的是ArrayList，不管啥方法，为了得到最终答案，移动数组的平均时间复杂度都要O(n)了。

代码

import java.util.ArrayList;

public class InsertInterval {
	public ArrayList<Interval> insert(ArrayList<Interval> intervals,
			Interval newInterval) {
		ArrayList<Interval> result = new ArrayList<Interval>();
		int N = intervals.size();
		for (int i = 0; i < N; ++i) {
			Interval interval = intervals.get(i);
			if (interval.start > newInterval.end) {
				result.add(newInterval);
				for (int j = i; j < N; ++j) {
					result.add(intervals.get(j));
				}
				return result;
			} else if (interval.end < newInterval.start) {
				result.add(interval);
			} else {
				newInterval.start = Math.min(newInterval.start, interval.start);
				newInterval.end = Math.max(newInterval.end, interval.end);
			}
		}
		result.add(newInterval);
		return result;
	}
}

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LeetCode | Insert Interval

原文：http://blog.csdn.net/perfect8886/article/details/21780789