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                  "
]

Swift解法

class Solution {
    func fullJustify(_ words: [String], _ maxWidth: Int) -> [String] {
        var i = 0
        var texts = [String]()
        var tempArray = [String]()
        var width = 0
        let spacingWidth = 1
        while i < words.count {
            let nextWidth = width + words[i].count + (tempArray.isEmpty ? 0 : spacingWidth)
            let isLastOne = (i == words.count - 1)
            var overflow = false
            if nextWidth <= maxWidth {
                width = nextWidth
                tempArray.append(words[i])
                i += 1
            } else {
                overflow = true
            }
            if isLastOne || overflow {
                let wordWidth = tempArray.map({$0.count}).reduce(0, {$0 + $1})
                let spacingCount = maxWidth - wordWidth
                var text = ""
                if (tempArray.count <= 1 || isLastOne) && !(overflow && tempArray.count > 1) {
                    text = tempArray.joined(separator: " ")
                    let remainSpacingCount = spacingCount - (tempArray.count - 1)
                    if remainSpacingCount > 0 {
                        text.append(String(repeating: " ", count: remainSpacingCount))
                    }
                } else {
                    let averageSpacingCount = spacingCount / (tempArray.count - 1)
                    var remainder = spacingCount % (tempArray.count - 1)
                    for string in tempArray {
                        if !text.isEmpty {
                            let length = averageSpacingCount + (remainder > 0 ? 1 : 0)
                            text.append(String(repeating: " ", count: length))
                            remainder -= 1
                        }
                        text.append(string)
                    }
                }
                texts.append(text)
                tempArray.removeAll()
                width = 0
            }
        }
        return texts
    }
}

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/text-justification