Given an array of words and a length L, format the text such that each line has exactly L characters and is fully (left and right) justified.
You should pack your words in a greedy approach; that is, pack as many words as you can in each line. Pad extra spaces ' '
when necessary so that each line has exactly Lcharacters.
Extra spaces between words should be distributed as evenly as possible. If the number of spaces on a line do not divide evenly between words, the empty slots on the left will be assigned more spaces than the slots on the right.
For the last line of text, it should be left justified and no extra space is inserted between words.
For example,
words: ["This", "is", "an", "example", "of", "text", "justification."]
L: 16
.
Return the formatted lines as:
[ "This is an", "example of text", "justification. " ]
Note: Each word is guaranteed not to exceed L in length.
- A line other than the last line might contain only one word. What should you do in this case?
In this case, that line should be left-justified.
public List<String> fullJustify(String[] words, int L) { List<String> list = new ArrayList<>(); int n = words.length; char[] spaces = new char[L]; Arrays.fill(spaces, ' '); for(int i=0; i<n; i++) { int j = i; int len = words[i].length(); while(i<n-1 && len+1+words[i+1].length()<=L) { len += 1+words[++i].length(); } //只有一个word或者是最后一行的时候,需要左对齐 boolean left = (j==i || i==n-1); int avg = left ? 0 : (L-len) / (i-j); //平均空格个数-1 int rem = left ? 0 : (L-len) % (i-j); //平均之后多出来的空格个数 StringBuilder sb = new StringBuilder(words[j]); while(j < i) { sb.append(spaces, 0, avg+1); if(rem-- > 0) sb.append(' '); sb.append(words[++j]); } sb.append(spaces, 0, L-sb.length()); list.add(sb.toString()); } return list; }
重构了一下Java代码:
public List<String> fullJustify(String[] words, int L) { List<String> res = new ArrayList<>(); int n = words.length; char[] spaces = new char[L]; Arrays.fill(spaces, ' '); for(int i=0; i<n; i++) { int len = words[i].length(); int j = i; while(i<n-1 && len+1+words[i+1].length()<=L) { len += 1+words[++i].length(); } StringBuilder sb = new StringBuilder(words[j]); if(j == i || i==n-1) { while(i==n-1 && j < i) { sb.append(" "+words[++j]); } sb.append(spaces, 0, L-sb.length()); } else { int avg = (L-len)/(i-j); int rem = (L-len)%(i-j); while(j < i) { sb.append(spaces, 0, avg+1); if(rem-- > 0) sb.append(" "); sb.append(words[++j]); } } res.add(sb.toString()); } return res; }
C++的代码:
vector<string> fullJustify(vector<string>& words, int maxWidth) { vector<string> result; int i = 0, n = words.size(); while(i < n) { int j(i), cnt(0), len(0); while(i<n && len+cnt+words[i].size()<=maxWidth) { len += words[i++].size(); cnt++; } bool left = (cnt==1 || i==n); // should be left aligned or not int space = left ? 1 : (maxWidth-len) / (cnt-1); int rem = left ? 0 : (maxWidth-len) % (cnt-1); string s; while(j < i) { s.append(words[j++]); if(j < i) s.append(space, ' '); if(rem > 0) { s.append(1, ' '); rem--; } } if(s.size() < maxWidth) { s.append(maxWidth-s.size(), ' '); } result.push_back(s); } return result; }
相关推荐
大佬的leetcode刷题笔记(c++版本)
leetcode-cli-plugins leetcode-cli 的第 3 方插件。 什么是 如何使用 如何使用 插件 名称 描述 增强的命令 按公司或标签过滤问题 list 不要在同一台计算机上使 Chrome 的会话过期 login 不要在同一台计算机上使 ...
LeetCode题解 - Java语言实现-181页.pdf
leetcode 答案Leetcode---数据库 我对 Leetcode 数据库问题的回答
彩色版本 正版 pdf 精讲数据结构 + 算法 链表 树 图表 贪心算法 指针 动态规划 查找算法
leetcode1-300.docx
leetcode 答案leetcode--python Leetcode 的答案
500195422331430LeetCode题解 - Java语言实现.zip
leetcode 答案LeetCode--哈希表 以上是我对“哈希表”问题的回答,都被leetcode的判断所接受。
Algorithm-LeetCode-Sol-Res.zip,干净,易懂的解决方案和资源,为leetcode在线判断算法问题。,算法是为计算机程序高效、彻底地完成任务而创建的一组详细的准则。
leetcode-cli 注意:这个存储库是为了临时使用而分叉的。 注意:从 webbrowser 复制 cookie 并使用leetcode user -c可以临时修复不能。 一个享受 leetcode 的高效 cli 工具! 非常感谢 leetcode.com,一个非常棒的...
leetcode 1 -200题所有源码 有问题私聊我
leetcode1-240题中文题解,md格式,java版本 有题目有题解有代码 需要使用markdown打开
leetcode26 algo 算法与数据结构,练习代码 语言:java 8 开发工具:vscode,安装插件Java Extension Pack vscode有智能提示,可调试,有重构支持,满足代码练习要求,相比IDEA更轻量级,普通笔记本即可流畅运行。 ...
leetcode 答案Leetcode---算法 我对 Leetcode 算法问题的回答
本书包含了 LeetCode Online Judge所有题目的答案,所有代码经过精心编写,编码规范良好,适合读者反复揣摩,模仿,甚至在纸上默写