[leetcode]56. Merge Intervals(Java)

https://leetcode.com/problems/merge-intervals/#/description

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].

Java Code:

package go.jacob.day626;import java.util.ArrayList;import java.util.Collections;import java.util.Comparator;import java.util.List;public class Demo3 {/* * Runtime: 25 ms.Your runtime beats 83.79 % of java submissions. * 参考自：@danchou  */public List<Interval> merge(List<Interval> intervals) {Collections.sort(intervals, new Comparator<Interval>() {@Overridepublic int compare(Interval obj0, Interval obj1) {return obj0.start - obj1.start;}});List<Interval> ret = new ArrayList<>();Interval prev = null;for (Interval inter : intervals) {if (prev == null || inter.start > prev.end) {ret.add(inter);prev = inter;} else if (inter.end > prev.end) {// Modify the element already in listprev.end = inter.end;}}return ret;}/* * 首先对intervals中的元素进行排序(按start升序) 然后循环判断能否增加end *  * Runtime: 27 ms.Your runtime beats 68.55 % of java submissions. */public List<Interval> merge_byme(List<Interval> intervals) {List<Interval> res = new ArrayList<Interval>();if (intervals == null)return res;// 对intervals中元素进行排序Collections.sort(intervals, new Comparator<Interval>() {@Overridepublic int compare(Interval i1, Interval i2) {return i1.start - i2.start;}});int index = 0;while (index < intervals.size()) {Interval temp1 = intervals.get(index);while (index != intervals.size() - 1) {Interval temp2 = intervals.get(index + 1);if (temp2.start <= temp1.end) {index++;if (temp2.end > temp1.end) {temp1.end = temp2.end;}} elsebreak;}res.add(temp1);index++;}return res;}}class Interval {int start;int end;Interval() {start = 0;end = 0;}Interval(int s, int e) {start = s;end = e;}}