LeetCode（24）文本左右对齐【数组/字符串】【困难】
目录
-
- 1.题目
- 2.答案
- 3.提交结果截图
链接： 文本左右对齐
1.题目
给定一个单词数组 words 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。
注意:
- 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 1:
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 "
]
提示:
1 <= words.length <= 3001 <= words[i].length <= 20words[i]由小写英文字母和符号组成1 <= maxWidth <= 100words[i].length <= maxWidth
2.答案
class Solution {
public List<String> fullJustify(String[] words, int maxWidth) {
if (words.length == 1) {
int leftSpaceNum = maxWidth - words[0].length();
char[] leftSpaces = new char[leftSpaceNum];
Arrays.fill(leftSpaces, ' ');
return Collections.singletonList(words[0] + new String(leftSpaces));
}
List<String> lines = new ArrayList<>();
int i = 1;
int length = words[0].length();
List<String> lineWords = new ArrayList<>();
lineWords.add(words[0]);
while (i < words.length) {
if (length + 1 + words[i].length() <= maxWidth) {
// 记录每行的单词
lineWords.add(words[i]);
length = length + 1 + words[i++].length();
} else {
// 已经记满一行
String line = wordsToLine(lineWords, maxWidth, false);
lines.add(line);
lineWords.clear();
lineWords.add(words[i]);
length = words[i++].length();
}
}
String line = wordsToLine(lineWords, maxWidth, true);
lines.add(line);
return lines;
}
/**
* 将单词转化为一行
* @param lineWords
* @return
*/
private String wordsToLine(List<String> lineWords, int maxWidth, boolean isLastLine) {
assert lineWords.size() >= 1;
int wordsLength = lineWords.stream().mapToInt(String::length).sum();
if (!isLastLine && lineWords.size() > 1) {
// 非最后一行
// 计算每个间隔空格
int eachSpaceNum = (maxWidth - wordsLength) / (lineWords.size() - 1);
char[] eachSpaces = new char[eachSpaceNum];
Arrays.fill(eachSpaces, ' ');
String eachSpaceStr = String.valueOf(eachSpaces);
// 计算第一个间隔额外空格数
int leftSpaceNum = (maxWidth - wordsLength) % (lineWords.size() - 1);
StringBuilder builder = new StringBuilder();
for (int i = 0; i < lineWords.size(); i++) {
builder.append(lineWords.get(i));
if (i != lineWords.size() - 1) {
builder.append(eachSpaceStr);
if (leftSpaceNum > 0) {
builder.append(" ");
leftSpaceNum--;
}
}
}
return builder.toString();
} else {
// 最后一行，左对齐即可
StringBuilder builder = new StringBuilder();
for (int i = 0; i < lineWords.size(); i++) {
builder.append(lineWords.get(i));
if (i != lineWords.size() - 1) {
builder.append(" ");
}
}
int leftSpaceNum = maxWidth - wordsLength - (lineWords.size() - 1);
char[] leftSpaces = new char[leftSpaceNum];
Arrays.fill(leftSpaces, ' ');
builder.append(new String(leftSpaces));
return builder.toString();
}
}
}
3.提交结果截图
整理完毕，完结撒花~