英文原文
Given an array of strings words and a width maxWidth, format the text such that each line has exactly maxWidth 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 maxWidth characters.
Extra spaces between words should be distributed as evenly as possible. If the number of spaces on a line does 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.
Note:
- A word is defined as a character sequence consisting of non-space characters only.
- Each word's length is guaranteed to be greater than 0 and not exceed maxWidth.
- The input array
wordscontains at least one word.
Example 1:
Input: words = ["This", "is", "an", "example", "of", "text", "justification."], maxWidth = 16 Output: [ "This is an", "example of text", "justification. " ]
Example 2:
Input: words = ["What","must","be","acknowledgment","shall","be"], maxWidth = 16 Output: [ "What must be", "acknowledgment ", "shall be " ] Explanation: Note that the last line is "shall be " instead of "shall be", because the last line must be left-justified instead of fully-justified. Note that the second line is also left-justified becase it contains only one word.
Example 3:
Input: words = ["Science","is","what","we","understand","well","enough","to","explain","to","a","computer.","Art","is","everything","else","we","do"], maxWidth = 20 Output: [ "Science is what we", "understand well", "enough to explain to", "a computer. Art is", "everything else we", "do " ]
Constraints:
1 <= words.length <= 3001 <= words[i].length <= 20words[i]consists of only English letters and symbols.1 <= maxWidth <= 100words[i].length <= maxWidth
中文题目
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 " ]
通过代码
高赞题解
思路
先取出一行能够容纳的单词，将这些单词根据规则填入一行
计算出额外空格的数量
spaceCount，额外空格就是正常书写用不到的空格- 由总长度算起
- 除去每个单词末尾必须的空格，为了统一处理可以在结尾虚拟加上一个长度
- 除去所有单词的长度
- 按照单词的间隙数量
wordCount - 1，对额外空格平均分布平均分布可查看 【另一篇题解】，简单来说就是商和余数的计算
- 对于每个词填充之后，需要填充的空格数量等于
spaceSuffix + spaceAvg + ((i - bg) < spaceExtra)spaceSuffix【单词尾部固定的空格】spaceAvg【额外空格的平均值，每个间隙都要填入 spaceAvg 个空格】((i - bg) < spaceExtra)【额外空格的余数，前 spaceExtra 个间隙需要多 1 个空格】
- 对于每个词填充之后，需要填充的空格数量等于
- 特殊处理
- 一行只有一个单词，单词左对齐，右侧填满空格
- 最后一行，所有单词左对齐，中间只有一个空格，最后一个单词右侧填满空格
答题
[]string fillWords(vector<string>& words, int bg, int ed, int maxWidth, bool lastLine = false) { int wordCount = ed - bg + 1; int spaceCount = maxWidth + 1 - wordCount; // 除去每个单词尾部空格， + 1 是最后一个单词的尾部空格的特殊处理 for (int i = bg; i <= ed; i++) { spaceCount -= words[i].size(); // 除去所有单词的长度 } int spaceSuffix = 1; // 词尾空格 int spaceAvg = (wordCount == 1) ? 1 : spaceCount / (wordCount - 1); // 额外空格的平均值 int spaceExtra = (wordCount == 1) ? 0 : spaceCount % (wordCount - 1); // 额外空格的余数 string ans; for (int i = bg; i < ed; i++) { ans += words[i]; // 填入单词 if (lastLine) // 特殊处理最后一行 { fill_n(back_inserter(ans), 1, ' '); continue; } fill_n(back_inserter(ans), spaceSuffix + spaceAvg + ((i - bg) < spaceExtra), ' '); // 根据计算结果补上空格 } ans += words[ed]; // 填入最后一个单词 fill_n(back_inserter(ans), maxWidth - ans.size(), ' '); // 补上这一行最后的空格 return ans; } vector<string> fullJustify(vector<string>& words, int maxWidth) { vector<string> ans; int cnt = 0; int bg = 0; for (int i = 0; i < words.size(); i++) { cnt += words[i].size() + 1; if (i + 1 == words.size() || cnt + words[i + 1].size() > maxWidth) { // 如果是最后一个单词，或者加上下一个词就超过长度了，即可凑成一行 ans.push_back(fillWords(words, bg, i, maxWidth, i + 1 == words.size())); bg = i + 1; cnt = 0; } } return ans; }
[]string fillWords(vector<string>& words, int bg, int ed, int maxWidth, bool lastLine = false) { int wordCount = ed - bg + 1; int spaceCount = maxWidth + 1 - wordCount; for (int i = bg; i <= ed; i++) { spaceCount -= words[i].size(); } int spaceSuffix = 1; int spaceAvg = (wordCount == 1) ? 1 : spaceCount / (wordCount - 1); int spaceExtra = (wordCount == 1) ? 0 : spaceCount % (wordCount - 1); string ans; for (int i = bg; i < ed; i++) { ans += words[i]; if (lastLine) { fill_n(back_inserter(ans), 1, ' '); continue; } fill_n(back_inserter(ans), spaceSuffix + spaceAvg + ((i - bg) < spaceExtra), ' '); } ans += words[ed]; fill_n(back_inserter(ans), maxWidth - ans.size(), ' '); return ans; } vector<string> fullJustify(vector<string>& words, int maxWidth) { vector<string> ans; int cnt = 0; int bg = 0; for (int i = 0; i < words.size(); i++) { cnt += words[i].size() + 1; if (i + 1 == words.size() || cnt + words[i + 1].size() > maxWidth) { ans.push_back(fillWords(words, bg, i, maxWidth, i + 1 == words.size())); bg = i + 1; cnt = 0; } } return ans; }
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