Leetcode 第68题：Text Justification-- 文本左右对齐（C++）

题目地址：Text Justification

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题目简介：

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。

需要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。

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:     
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        "
]
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。
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                  "
]

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题目解析：

参看地址：[LeetCode] Text Justification 文本左右对齐

C++版：

class Solution {
public:
    vector<string> fullJustify(vector<string> &words, int maxWidth) {
        vector<string> res;
        for (int i = 0; i < words.size();)
        {
            int j = i, len = 0;
            while (j < words.size() && len + words[j].size() + j - i <= maxWidth) 
            {
                len += words[j++].size();
            }
            string out;
            int space = maxWidth - len;
            for (int k = i; k < j; ++k) 
            {
                out += words[k];
                if (space > 0) 
                {
                    int tmp;
                    if (j == words.size())
                    { 
                        if (j - k == 1)    
                            tmp = space;
                        else 
                            tmp = 1;
                    } 
                    else
                    {
                        if (j - k - 1 > 0)
                        {
                            if (space % (j - k - 1) == 0) 
                                tmp = space / (j - k - 1);
                            else 
                                tmp = space / (j - k - 1) + 1;
                        } 
                        else 
                            tmp = space;
                    }
                    out.append(tmp, ' ');
                    space -= tmp;
                }
            }
            res.push_back(out);
            i = j;
        }
        return res;
    }
};