[LeetCode] 163. Missing Ranges 缺失区间
Given a sorted integer array nums, where the range of elements are in the inclusive range [lower, upper], return its missing ranges.
Example:
Input: nums =[0, 1, 3, 50, 75], lower = 0 and upper = 99,
Output:["2", "4->49", "51->74", "76->99"]
Java:
class Solution {
public List<String> findMissingRanges(int[] nums, int lower, int upper) {
List<String> res = new ArrayList<String>();
int next = lower;
for (int i = 0; i < nums.length; i++) {
// 1. We don't need to add [Integer.MAX_VALUE, ...] to result
if(lower == Integer.MAX_VALUE) return res;
if (nums[i] < next) {
continue;
}
if (nums[i] == next) {
next++;
continue;
}
res.add(getRange(next, nums[i] - 1));
// 2. We don't need to proceed after we have process Integer.MAX_VALUE in array
if(nums[i] == Integer.MAX_VALUE) return res;
next = nums[i] + 1;
}
if (next <= upper) {
res.add(getRange(next, upper));
}
return res;
}
public String getRange(int n1, int n2) {
return n1 == n2 ? String.valueOf(n1) : String.format("%d->%d" , n1, n2);
}
}
Java:
public class Solution {
public List<String> findMissingRanges(int[] A, int lower, int upper) {
if(A==null) return null;
List<String> res = new ArrayList<String>();
for(int i=0; i<A.length; i++) {
while(i<A.length && A[i] == lower ) {lower++; i++;}
if(i>=A.length) break;
if(A[i] == lower+1) {
res.add(String.valueOf(lower));
} else {
res.add("" + lower + "->" + (A[i]-1) );
}
lower = A[i] + 1;
}
if(lower == upper) {
res.add(String.valueOf(lower));
} else if(lower < upper ){
res.add("" + lower + "->" + upper );
}
return res;
}
}
Java:
public class Solution {
public List<String> findMissingRanges(int[] nums, int lower, int upper) {
List<String> res = new ArrayList<>();
if (nums == null || lower > upper) return res;
for (int num : nums) {
if (num - lower >= 1) res.add(getRangeString(lower, num - 1));
lower = num + 1;
}
if (lower <= upper) res.add(getRangeString(lower, upper));
return res;
}
private String getRangeString(int lower, int upper) {
if (lower == upper) return String.valueOf(lower);
StringBuilder sb = new StringBuilder();
sb.append(lower).append("->").append(upper);
return sb.toString();
}
}
Python:
class Solution(object):
def findMissingRanges(self, nums, lower, upper):
"""
:type nums: List[int]
:type lower: int
:type upper: int
:rtype: List[str]
"""
def getRange(lower, upper):
if lower == upper:
return "{}".format(lower)
else:
return "{}->{}".format(lower, upper)
ranges = []
pre = lower - 1 for i in xrange(len(nums) + 1):
if i == len(nums):
cur = upper + 1
else:
cur = nums[i]
if cur - pre >= 2:
ranges.append(getRange(pre + 1, cur - 1)) pre = cur return ranges
C++:
class Solution {
public:
vector<string> findMissingRanges(vector<int>& nums, int lower, int upper) {
vector<string> res;
int l = lower;
for (int i = 0; i <= nums.size(); ++i) {
int r = (i < nums.size() && nums[i] <= upper) ? nums[i] : upper + 1;
if (l == r) ++l;
else if (r > l) {
res.push_back(r - l == 1 ? to_string(l) : to_string(l) + "->" + to_string(r - 1));
l = r + 1;
}
}
return res;
}
};
类似题目:
[LeetCode] 228. Summary Ranges 总结区间
All LeetCode Questions List 题目汇总
[LeetCode] 163. Missing Ranges 缺失区间的更多相关文章
- [leetcode]163. Missing Ranges缺失范围
Given a sorted integer array nums, where the range of elements are in the inclusive range [lower, up ...
- LeetCode 163. Missing Ranges (缺失的区间)$
Given a sorted integer array where the range of elements are in the inclusive range [lower, upper], ...
- [LeetCode] Missing Ranges 缺失区间
Given a sorted integer array where the range of elements are [0, 99] inclusive, return its missing r ...
- ✡ leetcode 163. Missing Ranges 找出缺失范围 --------- java
Given a sorted integer array where the range of elements are in the inclusive range [lower, upper], ...
- [LeetCode#163] Missing Ranges
Problem: Given a sorted integer array where the range of elements are [lower, upper] inclusive, retu ...
- [LeetCode] 228. Summary Ranges 总结区间
Given a sorted integer array without duplicates, return the summary of its ranges. Example 1: Input: ...
- 【LeetCode】163. Missing Ranges 解题报告 (C++)
作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 遍历 日期 题目地址:https://leetcode ...
- 163. Missing Ranges
题目: Given a sorted integer array where the range of elements are [lower, upper] inclusive, return it ...
- 【LeetCode】Missing Ranges
Missing Ranges Given a sorted integer array where the range of elements are [lower, upper] inclusive ...
随机推荐
- xpath+多进程爬取八零电子书百合之恋分类下所有小说。
代码 # 需要的库 import requests from lxml import etree from multiprocessing import Pool import os # 请求头 he ...
- python测试开发django-rest-framework-64.序列化(serializers.Serializer)
前言 REST framework中的serializers与Django的Form和ModelForm类非常像.我们提供了一个Serializer类,它为你提供了强大的通用方法来控制响应的输出, 以 ...
- docker学习6-docker-compose容器集群编排
前言 实际工作中我们部署一个应用,一般不仅仅只有一个容器,可能会涉及到多个,比如用到数据库,中间件MQ,web前端和后端服务,等多个容器. 我们如果一个个去启动应用,当项目非常多时,就很难记住了,所有 ...
- *P2398 GCD SUM[数论]
题目描述 for i=1 to n for j=1 to n sum+=gcd(i,j) 解析 给出n求sum. gcd(x,y)表示x,y的最大公约数. 直接枚举复杂度为\(O(n^2)\),显然无 ...
- Alpha冲刺(9/10)——追光的人
1.队友信息 队员学号 队员博客 221600219 小墨 https://www.cnblogs.com/hengyumo/ 221600240 真·大能猫 https://www.cnblogs. ...
- less-5
首先输入id=1和id=1’未报错,均显示You are in.....(如下图所示) 由上图可以看到,如果运行返回结果正确的时候只返回you are in...,不会返回数据库当中的信息了,所以我们 ...
- Linux端口转发工具rinetd
介绍:Rinetd是为在一个Unix和Linux操作系统中为重定向传输控制协议(TCP)连接的一个工具.Rinetd是单一过程的服务器,它处理任何数量的连接到在配置文件etc/rinetd中指定的地址 ...
- Windows10 Faster R-CNN(GPU版) 配置训练自己的模型
参考链接 1. 找到合适自己的版本,下载安装Anaconda 点击跳转下载安装 Anaconda,双击下载好的 .exe 文件安装,只勾选第一个把 conda 添加到 PATH 路径.
- jsp大附件上传,支持断点续传
我们平时经常做的是上传文件,上传文件夹与上传文件类似,但也有一些不同之处,这次做了上传文件夹就记录下以备后用. 这次项目的需求: 支持大文件的上传和续传,要求续传支持所有浏览器,包括ie6,ie7,i ...
- mybatis-generator 插件
首先肯定要有mybatis的依赖 <!--mybatis spring--> <dependency> <groupId>org.mybatis.spring.bo ...