[LeetCode-interview algorithm classic-Java implementation] [057-Insert Interval (Insert Interval)], leetcode -- java

[LeetCode-interview algorithm classic-Java implementation] [057-Insert Interval (Insert Interval)], leetcode -- java
[057-Insert Interval (Insert Interval )][LeetCode-interview algorithm classic-Java implementation] [directory indexes for all questions]Original question

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[1,5],[6,9].
Example 2:
Given[1,2],[3,5],[6,7],[8,10],[12,16], Insert and merge[4,9]In[1,2],[3,10],[12,16].
This is because the new interval[4,9]Overlaps[3,5],[6,7],[8,10].

Theme

Inserts a new range for a series of non-covered intervals. If necessary, the intervals are merged starting from the start time.

Solutions

If the original interval is smaller than the inserted interval, insert the result set. If the inserted interval overlaps, update the inserted interval. If the inserted interval is smaller than the original interval, insert the inserted interval first, and then add the increased interval.

Code Implementation

Algorithm Implementation class

Import java. util. using list; import java. util. list; public class Solution {public List <Interval> insert (List <Interval> intervals, Interval newInterval) {// List of sets for saving results <Interval> result = new parameter List <> (); // if (intervals! = Null) {// traversal element for (Interval item: intervals) {// newInterval = null indicates that the inserted interval has been processed. // Add the interval smaller than the inserted interval to the result set if (newInterval = null | item. end <newInterval. start) {result. add (item) ;}// add an interval greater than the inserted interval to the result set, and add the inserted interval to the result set else if (item. start> newInterval. end) {result. add (newInterval); result. add (item); newInterval = null;} // The insert interval overlaps and the else {newInterval is updated. start = Math. min (newInterval. start, Item. start); newInterval. end = Math. max (newInterval. end, item. end) ;}}// if the inserted interval is not empty, the inserted interval is not processed. if (newInterval! = Null) {result. add (newInterval) ;}return result ;}}

Evaluation Result

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