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:

Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
Output: [[1,5],[6,9]]

Example 2:

Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
Output: [[1,2],[3,10],[12,16]]
Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].

Analysis

可以看成是Merge Intervals 的follow-up。这一问里non-overlapping, sorted的前提假设简直就是为了第一问准备的。

有了前一个问题的解法基础,那么这里的的区别就是在于何时插入新的interval。

因为排过序,用一个循环查找插入位置即可。时间复杂度:O(n)

插入操作java.util.ArrayList.add(),时间复杂度: O(n)

之后再进行Merge Intervals中完全一样的操作即可,时间复杂度:O(n)

空间复杂度,因为有额外的存储最终答案的ArrayList,空间复杂度为: O(n)

Solution

(13ms, 64%)

/**
 * Definition for an interval.
 * public class Interval {
 *     int start;
 *     int end;
 *     Interval() { start = 0; end = 0; }
 *     Interval(int s, int e) { start = s; end = e; }
 * }
 */
class Solution {
    public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
        List<Interval> ans = new ArrayList<>();

        int index = 0;
        while (index < intervals.size()) {
            if (newInterval.start > intervals.get(index).start) {
                index++;
            } else {
                break;
            }
        }
        intervals.add(index, newInterval);

        Interval last = null;
        for (Interval interval : intervals) {
            if (last == null || interval.start > last.end) {
                ans.add(interval);
                last = interval;
            } else {
                last.end = Math.max(last.end, interval.end);
            }
        }
        return ans;
    }
}

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