数据流合并成区间,每次新来一个数,表示成一个区间,然后在已经保存的区间中进行二分查找,最后结果有3种,插入头部,尾部,中间,插入头部,不管插入哪里,都判断一下左边和右边是否能和当前的数字接起来,我这样提交了,发现错了,想到之前考虑要不要判重,我感觉是这个问题,然后就是在二分查找的时候,判断一下左右区间是否包含当前的值,包含就直接返回。

 /**
* Definition for an interval.
* struct Interval {
* int start;
* int end;
* Interval() : start(0), end(0) {}
* Interval(int s, int e) : start(s), end(e) {}
* };
*/
bool cmp(Interval a, Interval b) {
if(a.start == b.start) return a.end < b.end;
return a.start < b.start;
}
class SummaryRanges {
public:
/** Initialize your data structure here. */ vector<Interval> v;
SummaryRanges() {
v.clear();
} void addNum(int val) {
//cout << val << " " << v.size() << endl;
Interval d(val, val);
if(v.size() == ) v.push_back(d);
else {
int t = lower_bound(v.begin(), v.end(), d, cmp) - v.begin();
//cout << val << " " << t << endl;
if(t == ) {
if(val == v[].start) return;
if(v[].start - == val) {
v[].start = val;
} else {
v.insert(v.begin() + t, d);
}
} else if(t == v.size()) {
if(v[t - ].end >= val) return;
if(v[t - ].end + == val) {
v[t - ].end = val;
} else {
v.push_back(d);
}
} else {
if(v[t - ].start == val || v[t - ].end >= val || v[t].start == val) return;
if(v[t - ].end + == v[t].start) {
v[t - ].end = v[t].end;
v.erase(v.begin() + t);
} else if(v[t - ].end + == val) {
v[t - ].end = val;
} else if(v[t].start - == val) {
v[t].start = val;
} else {
v.insert(v.begin() + t, d);
}
} }
} vector<Interval> getIntervals() {
return v;
}
}; /**
* Your SummaryRanges object will be instantiated and called as such:
* SummaryRanges obj = new SummaryRanges();
* obj.addNum(val);
* vector<Interval> param_2 = obj.getIntervals();
*/

[leetcode]352. Data Stream as Disjoint Intervals的更多相关文章

  1. [LeetCode] 352. Data Stream as Disjoint Intervals 分离区间的数据流

    Given a data stream input of non-negative integers a1, a2, ..., an, ..., summarize the numbers seen ...

  2. leetcode@ [352] Data Stream as Disjoint Intervals (Binary Search & TreeSet)

    https://leetcode.com/problems/data-stream-as-disjoint-intervals/ Given a data stream input of non-ne ...

  3. 【leetcode】352. Data Stream as Disjoint Intervals

    问题描述: Given a data stream input of non-negative integers a1, a2, ..., an, ..., summarize the numbers ...

  4. 352. Data Stream as Disjoint Intervals

    Plz take my miserable life T T. 和57 insert interval一样的,只不过insert好多. 可以直接用57的做法一个一个加,然后如果数据大的话,要用tree ...

  5. 352. Data Stream as Disjoint Intervals (TreeMap, lambda, heapq)

    Given a data stream input of non-negative integers a1, a2, ..., an, ..., summarize the numbers seen ...

  6. 352[LeetCode] Data Stream as Disjoint Intervals

    Given a data stream input of non-negative integers a1, a2, ..., an, ..., summarize the numbers seen ...

  7. [LeetCode] Data Stream as Disjoint Intervals 分离区间的数据流

    Given a data stream input of non-negative integers a1, a2, ..., an, ..., summarize the numbers seen ...

  8. Leetcode: Data Stream as Disjoint Intervals && Summary of TreeMap

    Given a data stream input of non-negative integers a1, a2, ..., an, ..., summarize the numbers seen ...

  9. [Swift]LeetCode352. 将数据流变为多个不相交间隔 | Data Stream as Disjoint Intervals

    Given a data stream input of non-negative integers a1, a2, ..., an, ..., summarize the numbers seen ...

随机推荐

  1. spring源码解析(一)---占位符解析替换

    一.结构类图 ①.PropertyResolver : Environment的顶层接口,主要提供属性检索和解析带占位符的文本.bean.xml配置中的所有占位符例如${}都由它解析 ②.Config ...

  2. centos 卸载自带的 java

    一般情况下,我们都要将linux自带的OPENJDK卸载掉,然后安装SUN的JDK 首先:查看Linux自带的JDK是否已安装     <1># java -version        ...

  3. Codeforces Round #320 (Div. 1) [Bayan Thanks-Round] C. Weakness and Poorness 三分 dp

    C. Weakness and Poorness Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/5 ...

  4. 理解class.forName()

    使用jdbc方式连接数据库时会使用一句代码Class.forName(String className).这句话是什么意思呢?首先说一点Class.forName(String className)这 ...

  5. Android中定时器的3种实现方法

    原文:http://blog.csdn.net/wulianghuan/article/details/8507221 在Android开发中,定时器一般有以下3种实现方法: 一.采用Handler与 ...

  6. android129 zhihuibeijing 聊天机器人

    上屏幕界面activity_main.xml: 语音识别界面 <LinearLayout xmlns:android="http://schemas.android.com/apk/r ...

  7. UNIX标准化及实现之选项

    POSIX.1的2001版,包括了ISO C标准所指定的各个函数.其接口分成了两类:必需接口和可选接口.可选接口按功能又进一步分成50个区.表1中按它们各自的选项代码总结了没有被弃用的编程接口.选项代 ...

  8. php开发环境配置 web UI模板

    web ui 能快速的整合进来?dwz? easyui?  bootstrap 在Apache 中配置: 在http.conf中加入php的设置 #php5_startphpIniDir " ...

  9. 10 Technologies That will Shape Future Education--reference

    http://dizyne.net/technologies-that-will-shape-future-education/ Technology is on the rise and with ...

  10. Android Studio无法启动 打开, Android Studio gradle下载不了

    Google在2013年I/O大会上发布了Android Studio,AndroidStudio是一个基于IntelliJ思想的新的Android开发工具.下面介绍一下Android Studio安 ...