228. Summary Ranges (everyday promlems) broken problems
Given a sorted integer array without duplicates, return the summary of its ranges.
Example 1:
Input: [0,1,2,4,5,7]
Output: ["0->2","4->5","7"]
Explanation: 0,1,2 form a continuous range; 4,5 form a continuous range.
Example 2:
Input: [0,2,3,4,6,8,9]
Output: ["0","2->4","6","8->9"]
Explanation: 2,3,4 form a continuous range; 8,9 form a continuous range.
broken solution : check the boundary
class Solution {
public List<String> summaryRanges(int[] nums) {
List<String> res = new ArrayList<>();
int n = nums.length;
if(n==0) return res;
else if(n==1) {
res.add(nums[0]+"");
return res;
}
int i = 0;
while(i<n-1){
//case for only n-1
if(nums[i] +1 == nums[i+1] ){ //
int start = nums[i];
i = i+1; while(nums[i] +1 == nums[i+1]){//i+1<n i++;
}
int end = nums[i];
String str = start + "->" + end;
res.add(str);
}
else if(nums[i] +1 != nums[i+1] ){
res.add(nums[i]+"");}
i++;
}
//check the last element
if(nums[n-1] == nums[n-2]+1) {
if(n>=3){
String[] temp = res.get(res.size()-1).split("->");
res.remove(res.size()-1);
res.add(temp[0] + "->" + nums[n-1]);
}else {
//n ==2
res.add(nums[n-2] + "->" + nums[n-1]);
}
}else {
res.add(nums[n-1]+"");
}
return res;
}
}
revision correct one: always check the boundary
two cases: 1: 1,2,3 2: [1,2,4,5,7]
class Solution {
public List<String> summaryRanges(int[] nums) {
List<String> res = new ArrayList<>();
int n = nums.length;
if(n==0) return res;
else if(n==1) {
res.add(nums[0]+"");
return res;
}
int i = 1;
while(i<=n-1){
//case for only n-1
if(nums[i-1] +1 == nums[i] ){ //
int start = nums[i-1];
i = i+1; while(i<n && nums[i-1] +1 == nums[i]){//i+1<n
i++;
}
int end = nums[i-1];
String str = start + "->" + end;
res.add(str);
}
else if(nums[i-1] +1 != nums[i] ){
res.add(nums[i-1]+"");}
i++;
}
//check the last element
if(nums[n-1] == nums[n-2]+1) { }else {
res.add(nums[n-1]+"");
}
return res;
}
}
two pointer solution
class Solution {
public List<String> summaryRanges(int[] nums) {
List<String> res = new ArrayList<>();
int n = nums.length;
if(n==0) return res;
else if(n==1) {
res.add(nums[0]+"");
return res;
}
//use two pointers
int i = 0;int j = 0;//i and j
while(i<n){
j = i+1;
while(j < n){
if(nums[j-1]+1 == nums[j]){
j++;
}else {
break;
}
}
if(j == i+1){
res.add(nums[i]+"");
}else {
res.add(nums[i]+"->"+nums[j-1]);
}
i = j;
}
return res;
}
}
ren ruoyousuopcheng, biypusuozhi
228. Summary Ranges (everyday promlems) broken problems的更多相关文章
- leetcode-【中等题】228. Summary Ranges
题目: 228. Summary Ranges Given a sorted integer array without duplicates, return the summary of its r ...
- 【LeetCode】228. Summary Ranges 解题报告(Python)
[LeetCode]228. Summary Ranges 解题报告(Python) 标签(空格分隔): LeetCode 题目地址:https://leetcode.com/problems/sum ...
- 【刷题-LeetCode】228. Summary Ranges
Summary Ranges Given a sorted integer array without duplicates, return the summary of its ranges. Ex ...
- Java for LeetCode 228 Summary Ranges
Given a sorted integer array without duplicates, return the summary of its ranges. For example, give ...
- LeetCode(228) Summary Ranges
Given a sorted integer array without duplicates, return the summary of its ranges. For example, give ...
- 228. Summary Ranges
Given a sorted integer array without duplicates, return the summary of its ranges. For example, give ...
- (easy)LeetCode 228.Summary Ranges
Given a sorted integer array without duplicates, return the summary of its ranges. For example, give ...
- 【LeetCode】228 - Summary Ranges
Given a sorted integer array without duplicates, return the summary of its ranges. For example, give ...
- Java [Leetcode 228]Summary Ranges
题目描述: Given a sorted integer array without duplicates, return the summary of its ranges. For example ...
随机推荐
- Codeforces 316C2 棋盘模型
Let's move from initial matrix to the bipartite graph. The matrix elements (i, j) for which i + j ar ...
- photoshop特效字体
一.3D效果字 3D效果文字给人以纵伸感.立体感和真实感,是商家常用到的一种宣传文字.虽然Photoshop软件是平面软件,但是在制作3D效果文字时却游刃有余. 3D效果字的制作可分以下三步完成. 输 ...
- Java中Array与ArrayList的主要区别
1)精辟阐述: 可以将 ArrayList想象成一种"会自动扩增容量的Array". 2)Array([]):最高效:但是其容量固定且无法动态改变: ArrayList: ...
- Pycharm在线/手动离线安装第三方库-以scapy为例(本地离线添加已经安装的第三方库通过添加Path实现)
在线安装运行Pycharm,打开需要添加scapy文件的项目,以TestScapy为例 点击工具栏的File选项,选中Settings,单击打开 ...
- 第五章:引用类型(一)-Object和Array
引用类型 引用类型的值(对象)是引用类型的一个实例 引用类型是一种数据结构,用于将数据与功能组织在一起 也常被称为类, Object 对象的两种创建方式 使用new操作符 对象字面量表示法 Array ...
- 【Java】Java中的Collections类——Java中升级版的数据结构【转】
一般来说课本上的数据结构包括数组.单链表.堆栈.树.图.我这里所指的数据结构,是一个怎么表示一个对象的问题,有时候,单单一个变量声明不堪大用,比如int,String,double甚至一维数组.二维数 ...
- spring webapp的配置文件放置在项目外的方法
在web.xml中,填写 <context-param> <param-name>CFG_HOME</param-name> ...
- Unity Transform
public class PlayerControll : MonoBehaviour { Transform playerTransform; Animation playerAnimation; ...
- Hash表的原理
哈希的概念:Hash,一般翻译做“散列”,也有直接音译为“哈希”的,就是把任意长度的输入(又叫做预映射, pre-image),通过散列算法,变换成固定长度的输出,该输出就是散列值.这种转换是一种压缩 ...
- css有关鼠标移动上去图片变透明度变化
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/ ...