标签:interval binarysearch 二分查找 leetcode
【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]
.
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; } }
【LeetCode】Insert Interval 解题报告
标签:interval binarysearch 二分查找 leetcode
原文地址:http://blog.csdn.net/ljiabin/article/details/44923157