vector<Interval> merge(vector<Interval> &intervals) { vector<Interval> res; if(intervals.size()==0) return res; Interval temp; int flag=0;//推断上一次是否有区间合并 sort(intervals.begin(),intervals.end(),cmpLess()); for(
1/**2* Definition for an interval.3* public class Interval {4* int start;5* int end;6* Interval() { start = 0; end = 0; }7* Interval(int s, int e) { start = s; end = e; }8* }9*/10publicclassSolution11{12publicList<Interval> merge(List<Interval>intervals)13{14List<Int...
def merge(self, intervals): """ :type intervals: List[Interval] :rtype: List[Interval] """ if not intervals: return [] intervals.sort(key = lambda x:x.start) length=len(intervals) res=[intervals[0]] for i in range(1, length): size=len(res) if res[size-1].start<=intervals[i...
public List<Interval> merge2(List<Interval> intervals) { if(intervals == null || intervals.size() < 2) return intervals; List<Interval> res = new ArrayList<>(); int len = intervals.size(); int[] starts = new int[len], ends = new int[len]; for(int i = 0; i < len; i++)...
Can you solve this real interview question? Merge Intervals - Given an array of intervals where intervals[i] = [starti, endi], merge all overlapping intervals, and return an array of the non-overlapping intervals that cover all the intervals in the input
1/**2* Definition for an interval.3* public class Interval {4* int start;5* int end;6* Interval() { start = 0; end = 0; }7* Interval(int s, int e) { start = s; end = e; }8* }9*/10publicclassSolution {11publicList<Interval> merge(List<Interval>intervals) {12List<Inter...
Can you solve this real interview question? Merge Intervals - Given an array of intervals where intervals[i] = [starti, endi], merge all overlapping intervals, and return an array of the non-overlapping intervals that cover all the intervals in the input
56. 合并区间 - 以数组 intervals 表示若干个区间的集合,其中单个区间为 intervals[i] = [starti, endi] 。请你合并所有重叠的区间,并返回 一个不重叠的区间数组,该数组需恰好覆盖输入中的所有区间 。 示例 1: 输入:intervals = [[1,3],[2,6],[8,10],[15,18]] 输
class Solution { public int[][] merge(int[][] intervals) { if(intervals.length == 0) return new int[0][]; Arrays.sort(intervals, new Comparator<int[]>(){ public int compare(int[] interval1, int[] interval2){ return interval1[0] - interval2[0]; } }); List<int[]> merged ...
思路:先sort by start,然后不断判断当前的interval能不能吞噬下一个(curr.end >= next.start,说明这两个能连起来)。做的过程中发现几个问题: 本来是想直接循环list,然后不断吞噬,删掉被吞噬的。但是这样就一边iterate一边modify array了,会挂。所以还得用另一个list,然后决定是不断删list2里的东西,还是不断...