Leetcode #57. Insert Interval 间隔插入 解题报告

1 解题思想

首先，请先看昨天的基友题 Leetcode #56. Merge Intervals 间隔合并 解题报告

今天的题目也是给了一堆区间，然后再给一个额外的区间，让你将他进行插入，然后合并输出。

我的方法很懒，直接用昨天的代码就可以了。。。昨天的是合并啊，今天我们就把这个新的区间，加进去合并下不就可以了？

所以请仔细看看代码注释，或者是昨天的解题。。我真的复制粘贴后几十秒就搞定了

2 原题

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].

3 AC解

/**
* 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> merge(List<Interval> intervals) {
List<Interval> result=new ArrayList<Interval>();
Collections.sort(intervals, new Comparator<Interval>() {
public int compare(Interval arg0, Interval arg1) {
return arg0.start-arg1.start;
}
});
if(intervals.size()<1)
return result;
Interval p=intervals.get(0),tmp;
int i=1;
while(i<intervals.size()){
tmp=intervals.get(i);
if(p.end>=tmp.start){
p.end=Math.max(p.end,tmp.end);
}
else{
result.add(p);
p=tmp;
}
i++;
}
result.add(p);
return result;
}
/**
* 这道题直接复制就好了 优化点可以考虑下用二分查找顺序，不用再排序了。。但我偷懒
* 这有这两行代码是新写的，剩下复制的。。。30S AC https://leetcode.com/problems/merge-intervals/ 复制的是这题的代码

*/
public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
intervals.add(newInterval);
return merge(intervals);
}
}