文本左右对齐

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。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/text-justification
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

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 "
]

实现代码

class Solution {
public:
    vectorfullJustify(vector& words, int maxWidth)
    {
        vector res;
        mapwordsSqe;
        int n = 0;
        int length = 0;
        int step = 0;
        vector temp;
        while (n < words.size())
        {
            length += words[n].size();
            if (length < maxWidth)
            {
                temp.push_back(words[n]);
                length++;
            }
            else if (length == maxWidth)
            {
                temp.push_back(words[n]);
                res.push_back(temp);
                length = 0;
                temp.clear();
            }
            else if (length > maxWidth)
            {
                res.push_back(temp);
                temp.clear();
                temp.push_back(words[n]);
                length = words[n].size();
                length++;
            }
            n++;
        }
        if (!temp.empty())
            res.push_back(temp);
        vector result;
        for (int i = 0; i < res.size(); i++)
        {
            string tempString;
            int totalLength = 0;
            for (int j = 0; j < res[i].size(); j++)
                totalLength += res[i][j].size();
            int nullNumber;
            if (i != res.size() - 1)
            {
                if (res[i].size() > 1)
                {
                    nullNumber = (maxWidth - totalLength) / (res[i].size() - 1);
                    int fullNumber = (maxWidth - totalLength) % (res[i].size() - 1);
                    int count = 0;
                    for (int j = 0; j < res[i].size(); j++)
                    {
                        if (j != res[i].size() - 1)
                        {
                            if (fullNumber == 0)
                                tempString.append(res[i][j]).append(nullNumber, ' ');
                            else
                            {
                                if (count != fullNumber)
                                {
                                    tempString.append(res[i][j]).append(nullNumber + 1, ' ');
                                    count++;
                                }
                                else
                                    tempString.append(res[i][j]).append(nullNumber, ' ');
                            }
                        }
                        else
                            tempString.append(res[i][j]);
                    }
                }
                else
                    tempString.append(res[i][0]).append(maxWidth - totalLength, ' ');
            }
            else
            {
                if (res[i].size() > 1)
                {
                    nullNumber = maxWidth - totalLength - (res[i].size() - 1);
                    for (int j = 0; j < res[i].size(); j++)
                    {
                        if (j != res[i].size() - 1)
                            tempString.append(res[i][j]).append(" ");
                        else
                            tempString.append(res[i][j]).append(nullNumber, ' ');
                    }
                }
                else
                    tempString.append(res[i][0]).append(maxWidth - totalLength, ' ');
            }
            result.push_back(tempString);
        }
        return result;
    }
};