68:文本左右对齐

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huangge1199 2021-08-30 16:43:59 +08:00
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//给定一个单词数组和一个长度 maxWidth重新排版单词使其成为每行恰好有 maxWidth 个字符且左右两端对齐的文本
//
// 你应该使用贪心算法来放置给定的单词也就是说尽可能多地往每行中放置单词必要时可用空格 ' ' 填充使得每行恰好有 maxWidth 个字符
//
// 要求尽可能均匀分配单词间的空格数量如果某一行单词间的空格不能均匀分配则左侧放置的空格数要多于右侧的空格数
//
// 文本的最后一行应为左对齐且单词之间不插入额外的空格
//
// 说明:
//
//
// 单词是指由非空格字符组成的字符序列
// 每个单词的长度大于 0小于等于 maxWidth
// 输入单词数组 words 至少包含一个单词
//
//
// 示例:
//
// 输入:
//words = ["This", "is", "an", "example", "of", "text", "justification."]
//maxWidth = 16
//输出:
//[
//   "This    is    an",
//   "example  of text",
//   "justification.  "
//]
//
//
// 示例 2:
//
// 输入:
//words = ["What","must","be","acknowledgment","shall","be"]
//maxWidth = 16
//输出:
//[
//  "What   must   be",
//  "acknowledgment  ",
//  "shall be        "
//]
//解释: 注意最后一行的格式应为 "shall be " 而不是 "shall be",
//  因为最后一行应为左对齐而不是左右两端对齐
// 第二行同样为左对齐这是因为这行只包含一个单词
//
//
// 示例 3:
//
// 输入:
//words = ["Science","is","what","we","understand","well","enough","to",
//"explain",
//  "to","a","computer.","Art","is","everything","else","we","do"]
//maxWidth = 20
//输出:
//[
//  "Science  is  what we",
// "understand      well",
//  "enough to explain to",
//  "a  computer.  Art is",
//  "everything  else  we",
//  "do                  "
//]
//
// Related Topics 字符串 模拟 👍 149 👎 0
package leetcode.editor.cn;
import java.util.ArrayList;
import java.util.List;
//68:文本左右对齐
class TextJustification {
public static void main(String[] args) {
//测试代码
Solution solution = new TextJustification().new Solution();
System.out.println(solution.fullJustify(new String[]{"This", "is", "an", "example", "of",
"text", "justification."}, 16));
}
//力扣代码
//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
public List<String> fullJustify(String[] words, int maxWidth) {
List<String> result = new ArrayList<>();
List<String> lines = new ArrayList<>();
int size = 0;
for (String word : words) {
if (size + word.length() > maxWidth) {
StringBuilder line = new StringBuilder();
if (lines.size() == 1) {
line.append(lines.get(0));
int temp = maxWidth - line.length();
for (int j = 0; j < temp; j++) {
line.append(" ");
}
} else {
int dev = (maxWidth - size + 1) / (lines.size() - 1);
int mod = (maxWidth - size + 1) % (lines.size() - 1);
StringBuilder space = new StringBuilder();
for (int j = 0; j <= dev; j++) {
space.append(" ");
}
for (int j = 0; j < lines.size() - 1; j++) {
String str = lines.get(j);
if (j < mod) {
str += space + " ";
} else {
str += space;
}
line.append(str);
}
line.append(lines.get(lines.size() - 1));
}
result.add(line.toString());
lines = new ArrayList<>();
size = 0;
}
lines.add(word);
size += word.length() + 1;
}
StringBuilder str = new StringBuilder();
for (String line : lines) {
str.append(line).append(" ");
}
str = new StringBuilder(str.substring(0, str.length() - 1));
size = maxWidth - str.length();
for (int i = 0; i < size; i++) {
str.append(" ");
}
result.add(str.toString());
return result;
}
}
//leetcode submit region end(Prohibit modification and deletion)
}

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<p>给定一个单词数组和一个长度&nbsp;<em>maxWidth</em>,重新排版单词,使其成为每行恰好有&nbsp;<em>maxWidth</em>&nbsp;个字符,且左右两端对齐的文本。</p>
<p>你应该使用&ldquo;贪心算法&rdquo;来放置给定的单词;也就是说,尽可能多地往每行中放置单词。必要时可用空格&nbsp;<code>&#39; &#39;</code>&nbsp;填充,使得每行恰好有 <em>maxWidth</em>&nbsp;个字符。</p>
<p>要求尽可能均匀分配单词间的空格数量。如果某一行单词间的空格不能均匀分配,则左侧放置的空格数要多于右侧的空格数。</p>
<p>文本的最后一行应为左对齐,且单词之间不插入<strong>额外的</strong>空格。</p>
<p><strong>说明:</strong></p>
<ul>
<li>单词是指由非空格字符组成的字符序列。</li>
<li>每个单词的长度大于 0小于等于&nbsp;<em>maxWidth</em></li>
<li>输入单词数组 <code>words</code>&nbsp;至少包含一个单词。</li>
</ul>
<p><strong>示例:</strong></p>
<pre><strong>输入:</strong>
words = [&quot;This&quot;, &quot;is&quot;, &quot;an&quot;, &quot;example&quot;, &quot;of&quot;, &quot;text&quot;, &quot;justification.&quot;]
maxWidth = 16
<strong>输出:</strong>
[
&nbsp; &nbsp;&quot;This &nbsp; &nbsp;is &nbsp; &nbsp;an&quot;,
&nbsp; &nbsp;&quot;example &nbsp;of text&quot;,
&nbsp; &nbsp;&quot;justification. &nbsp;&quot;
]
</pre>
<p><strong>示例&nbsp;2:</strong></p>
<pre><strong>输入:</strong>
words = [&quot;What&quot;,&quot;must&quot;,&quot;be&quot;,&quot;acknowledgment&quot;,&quot;shall&quot;,&quot;be&quot;]
maxWidth = 16
<strong>输出:</strong>
[
&nbsp; &quot;What &nbsp; must &nbsp; be&quot;,
&nbsp; &quot;acknowledgment &nbsp;&quot;,
&nbsp; &quot;shall be &nbsp; &nbsp; &nbsp; &nbsp;&quot;
]
<strong>解释: </strong>注意最后一行的格式应为 &quot;shall be &quot; 而不是 &quot;shall be&quot;,
&nbsp; 因为最后一行应为左对齐,而不是左右两端对齐。
第二行同样为左对齐,这是因为这行只包含一个单词。
</pre>
<p><strong>示例&nbsp;3:</strong></p>
<pre><strong>输入:</strong>
words = [&quot;Science&quot;,&quot;is&quot;,&quot;what&quot;,&quot;we&quot;,&quot;understand&quot;,&quot;well&quot;,&quot;enough&quot;,&quot;to&quot;,&quot;explain&quot;,
&nbsp; &quot;to&quot;,&quot;a&quot;,&quot;computer.&quot;,&quot;Art&quot;,&quot;is&quot;,&quot;everything&quot;,&quot;else&quot;,&quot;we&quot;,&quot;do&quot;]
maxWidth = 20
<strong>输出:</strong>
[
&nbsp; &quot;Science &nbsp;is &nbsp;what we&quot;,
&quot;understand &nbsp; &nbsp; &nbsp;well&quot;,
&nbsp; &quot;enough to explain to&quot;,
&nbsp; &quot;a &nbsp;computer. &nbsp;Art is&quot;,
&nbsp; &quot;everything &nbsp;else &nbsp;we&quot;,
&nbsp; &quot;do &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&quot;
]
</pre>
<div><div>Related Topics</div><div><li>字符串</li><li>模拟</li></div></div><br><div><li>👍 149</li><li>👎 0</li></div>