[LeetCode-interview algorithm classic-Java implementation] [056-Merge Intervals (range Merge)], leetcode -- java

[LeetCode-interview algorithm classic-Java implementation] [056-Merge Intervals (range Merge)], leetcode -- java
[056-Merge Intervals (Merge Intervals )][LeetCode-interview algorithm classic-Java implementation] [directory indexes for all questions]Original question

Given a collection of intervals, merge all overlapping intervals.
For example,
Given[1,3],[2,6],[8,10],[15,18],
Return[1,6],[8,10],[15,18].

Theme

Given a set of intervals, merge overlapping intervals.

Solutions

Sort the intervals first, sort by start point, and then merge them one by one.

Code Implementation

Algorithm Implementation class

Import java. util. *; public class Solution {public List <Interval> merge (List <Interval> intervals) {List <Interval> result = new partition List <> (); if (intervals = null | intervals. size () <1) {return result;} // sort the interval first and use an anonymous internal class Collections. sort (intervals, new Comparator <Interval> () {@ Override public int compare (Interval o1, Interval o2) {return o1.start-o2.start ;}}); // after sorting, the start of the last element (marked as next) must be no less than the previous (marked as prev) start, // For the newly added interval, if next. start is greater than prev. the end indicates that the two intervals are separated, and a new interval needs to be added //; otherwise, next. start in [prev. start, prev. end], you only need to read // next. whether end is greater than prev. end. if the value is greater than the value, you need to merge the range (extended) Interval prev = null; for (Interval item: intervals) {if (prev = null | prev. end <item. start) {result. add (item); prev = item;} else if (prev. end <item. end) {prev. end = item. end ;}} return result ;}}

Evaluation Result

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