题目来源：https://leetcode.com/problems/insert-interval/

问题描述

57. Insert Interval

Hard

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].

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题意

给定一个由Interval（由start和end）组成的有序数列intervals，将新的Interval（newInterval）插入其中（可能需要merge），返回插入后的intervals

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思路

首先二分查找查找第一个始端>=newInterval的始端的位置ind, 向前看一个interval看是否发生重叠，向后while循环看若干个Interval可能重叠。删去重叠的Interval, 插入merge过后的Interval. 具体实现的时候注意remove之后数列中的被remove元素之后的元素的下标都会发生变化。

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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; }
 * }
 */
class Solution {
    public int binarySearch(List<Interval> intervals, Interval newInterval)
    {
        int n = intervals.size(), l = 0, r = n-1, m = 0, s = newInterval.start;
        while (l <= r)
        {
            m = (l + r) / 2;
            if (intervals.get(m).start < s)
            {
                l = m + 1;
            }
            else
            {
                r = m - 1;
            }
        }
        return l;
    }
    
    public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
        int n = intervals.size(), e = newInterval.end;
        if (n == 0)
        {
            intervals.add(newInterval);
            return intervals;
        }
        int merge_begin = binarySearch(intervals, newInterval), i = merge_begin, val_end = 0;
        if (merge_begin == n)
        {
            if (intervals.get(n-1).end >= newInterval.start)
            {
                Interval removed = intervals.remove(n-1);
                intervals.add(new Interval(removed.start, Math.max(e, removed.end)));
            }
            else
            {
                intervals.add(newInterval);
            }
        }
        else
        {
            while (i < n && e >= intervals.get(i).start)
            {
                i++;
            }
            if (i > merge_begin)
            {
                val_end = Math.max(e, intervals.get(i-1).end);
                for (int j = merge_begin; j < i; j++)
                {
                    intervals.remove(merge_begin);
                }
            }
            if (merge_begin > 0 && intervals.get(merge_begin-1).end >= newInterval.start)
            {
                Interval removed = intervals.remove(merge_begin-1);
                if (i > merge_begin)
                {
                    intervals.add(merge_begin-1, new Interval(removed.start, val_end));
                }
                else
                {
                    intervals.add(merge_begin-1, new Interval(removed.start, Math.max(newInterval.end, removed.end)));
                }
            }
            else
            {
                if (i > merge_begin)
                {
                    intervals.add(merge_begin, new Interval(newInterval.start, val_end));
                }
                else
                {
                    intervals.add(merge_begin, new Interval(newInterval.start, newInterval.end));
                }
            }
        }
        return intervals;
    }
}

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