leetcode 57. Insert Interval

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本文介绍了一个高效的算法，用于在一个已排序的区间集合中插入一个新的区间，并在必要时进行合并。通过示例说明了如何实现这一过程，确保了新插入的区间能够正确地与现有区间合并。

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].

和上一题基本相同，不过本身是有序的，需要插入一个新的interval

思路是先把end<new.start的interval先放入，找到第一个end>=new.start的interval，然后在找到第一个start>new.end的interval，放入new，

再把剩余的interval逐个放入

/**
 * 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> res = new ArrayList<Interval>();
        if(intervals.size()<1){
            res.add(newInterval);
            return res;
        }
        int i = 0;
        while(i<intervals.size()&&intervals.get(i).end<newInterval.start){
            res.add(intervals.get(i));
            i++;
        }
        if(i==intervals.size()){
            res.add(newInterval);
            return res;
        }
        newInterval.start = Math.min(newInterval.start,intervals.get(i).start);
        while(i<intervals.size() && intervals.get(i).start<=newInterval.end)  
        {  
            newInterval.end = Math.max(newInterval.end, intervals.get(i).end);  
            i++;  
        }
        res.add(newInterval);
        while(i<intervals.size())  
        {  
            res.add(intervals.get(i));  
            i++;  
        }  
    return res;
    }
}