Leetcode 57. 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].

思路

思路一:将这个问题转化成56题来做。时间开销稍微大一点，因为做了不必要的计算
思路二:先找跟newinterval相关的几个interval，然后将这几个合并，然后跟不相干的几个在合并(未实现)

测试用例

[1,3]6,9] [2,5]
[1,2][2,4] []
[] [1,2]

代码

/** * 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; } * } */public class Solution {    public List<Interval> insert(List<Interval> intervals, Interval newInterval) {        intervals.add(newInterval);        return merge(intervals);    }    public List<Interval> merge(List<Interval> intervals) {        // sort start&end        int n = intervals.size();        int[] starts = new int[n];        int[] ends = new int[n];        for (int i = 0; i < n; i++) {            starts[i] = intervals.get(i).start;            ends[i] = intervals.get(i).end;        }        Arrays.sort(starts);        Arrays.sort(ends);        // loop through        List<Interval> res = new ArrayList<Interval>();        for (int i = 0, j = 0; i < n; i++) { // j is start of interval.            if (i == n - 1 || starts[i + 1] > ends[i]) {                res.add(new Interval(starts[j], ends[i]));                j = i + 1;            }        }        return res;    }}

他山之玉

public List<Interval> insert(List<Interval> intervals, Interval newInterval) {    List<Interval> result = new LinkedList<>();    int i = 0;    // add all the intervals ending before newInterval starts    while (i < intervals.size() && intervals.get(i).end < newInterval.start)        result.add(intervals.get(i++));    // merge all overlapping intervals to one considering newInterval    while (i < intervals.size() && intervals.get(i).start <= newInterval.end) {        newInterval = new Interval( // we could mutate newInterval here also                Math.min(newInterval.start, intervals.get(i).start),                Math.max(newInterval.end, intervals.get(i).end));        i++;    }    result.add(newInterval); // add the union of intervals we got    // add all the rest    while (i < intervals.size()) result.add(intervals.get(i++));     return result;}