leetCode 56.Merge Intervals (合并区间) 解题思路和方法

Merge Intervals

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].

思路：题意很明确，首先对各区间按开始来排序，最后遍历，如果前面和后面的区间有重合，合并。

具体代码：

/**
* 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> merge(List<Interval> intervals) {
List<Interval> list = new ArrayList<Interval>();
//排序，实现了Comparator接口
Collections.sort(intervals,new Comparator<Interval>() {

@Override
public int compare(Interval o1, Interval o2) {
// TODO Auto-generated method stub
return o1.start - o2.start;//起始值升序排序
}
});

if(intervals.size() == 0)
return list;

Interval i1 = intervals.get(0);
//遍历
for(int i = 0; i < intervals.size(); i++){
Interval i2;
//分情况i2赋值
if(i == intervals.size() - 1)//如果i到最后，增加一个虚拟最大的区间
i2 = new Interval(Integer.MAX_VALUE, Integer.MAX_VALUE);
else//否则，i2最后i1后面的值
i2 = intervals.get(i+1);
//合并区间
if(i2.start >= i1.start && i2.start <= i1.end){
i1.end = Math.max(i1.end, i2.end);
}else{//没有交集，直接添加
list.add(i1);
i1 = i2;//i1更迭
}
}
return list;
}
}