【LeetCode】Insert Interval 解题报告

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【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].

【解析】

题意：给一些已经按开始时间排好序的区间，现在往里面插入一个区间，如果有重叠就合并区间，返回插入后的区间序列。

思路：先插入，在合并重叠区间。既然是已经排好序的，就可以用二分查找的方法，把要插入的这个区间放到应该的位置。合并重叠区间的方法《【LeetCode】Merge Intervals 解题报告》一样。

【Java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) {        List<Interval> ans = new ArrayList<Interval>();                // insert newInterval by binary searching        int l = 0;        int r = intervals.size() - 1;        while (l <= r) {            int mid = (l + r) >> 1;            if (intervals.get(mid).start > newInterval.start) {                r = mid - 1;            } else {                l = mid + 1;            }        }        intervals.add(l, newInterval);                // merge all overlapping intervals        int start = intervals.get(0).start;        int end = intervals.get(0).end;        for (int i = 1; i < intervals.size(); i++) {            Interval inter = intervals.get(i);            if (inter.start > end) {                ans.add(new Interval(start, end));                start = inter.start;                end = inter.end;            } else {                end = Math.max(end, inter.end);            }        }        ans.add(new Interval(start, end));                return ans;    }}

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