Description There is given the series of n closed intervals [ai; bi], where i=1,2,...,n. The sum of those intervals may be represented as a sum of closed pairwise non−intersecting intervals. The task is to find such representation with the minimal nu…

Given a collection of intervals, find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping. Note: You may assume the interval's end point is always bigger than its start point. Intervals like [1,2] and…

Given a data stream input of non-negative integers a1, a2, ..., an, ..., summarize the numbers seen so far as a list of disjoint intervals. For example, suppose the integers from the data stream are 1, 3, 7, 2, 6, ..., then the summary will be: [1, 1…

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]. 这道和之前那道Insert Interval 插入区间 很类似,这次题目要求我们合并区间,之前那题明确了输入区间集是有序的,而这题没有,所有我们首先要做的就是给区间集排序,由于我们要排序的是个结构体,所以我们要定义自…

Intervals Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 26028   Accepted: 9952 Description You are given n closed, integer intervals [ai, bi] and n integers c1, ..., cn. Write a program that: reads the number of intervals, their end po…

Source: Sigma Zone, by Philip Mayfield The Binomial Distribution is commonly used in statistics in a variety of applications. Binomial data and statistics are presented to us daily. For example, in the election of political officials we may be asked…

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]. 题目的意思是将相交得区间合并,典型的贪心算法 首先将区间先按照start进行排序, 然后保存先前区间的start和end 如果当前的start > 先前的end,说明当前的区间与之前的区间不想交,则将先前的区间放入结果中…

感觉有一点进步了,但是思路还是不够犀利. /** * Definition for an interval. * struct Interval { * int start; * int end; * Interval() : start(0), end(0) {} * Interval(int s, int e) : start(s), end(e) {} * }; */ class Solution { public: vector<Interval> merge(vector<In…

class Solution { public: static bool cmp(Interval &a,Interval &b) { return a.start<b.start; } vector<Interval> merge(vector<Interval>& intervals) { vector<Interval> res; )return res; sort(intervals.begin(),intervals.end(),…

Given a collection of intervals, find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping. Note: You may assume the interval's end point is always bigger than its start point. Intervals like [1,2] and…