Leetcode 第68題：Text Justification-- 文本左右對齊（C++）

題目地址：Text Justification

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題目簡介：

給定一個單詞數組和一個長度 maxWidth，重新排版單詞，使其成爲每行恰好有 maxWidth 個字符，且左右兩端對齊的文本。

需要儘可能多地往每行中放置單詞，必要時可用空格 ' ' 填充，使得每行恰好有 maxWidth 個字符。

儘可能均勻分配單詞間的空格數量，如果某一行單詞間的空格不能均勻分配，則左側放置的空格數要多於右側的空格數。

文本的最後一行應爲左對齊，且單詞之間不插入額外的空格。說明:單詞是指由非空格字符組成的字符序列。每個單詞的長度大於 0，小於等於 maxWidth。

輸入單詞數組 words 至少包含一個單詞。

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        "
]
解釋: 注意最後一行的格式應爲 "shall be  " 而不是 "shall   be",因爲最後一行應爲左對齊，而不是左右兩端對齊。第二行同樣爲左對齊，這是因爲這行只包含一個單詞。
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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題目解析：

參看地址：[LeetCode] Text Justification 文本左右對齊

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;
    }
};