题目说明

给定一个单词数组 words 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。

注意:

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

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                  "
]

提示:

1 <= words.length <= 300
1 <= words[i].length <= 20
words[i] 由小写英文字母和符号组成
1 <= maxWidth <= 100
words[i].length <= maxWidth

代码实现

class Solution {
public:
    vector<string> fullJustify(vector<string>& words, int maxWidth) {
        vector<string> res;
        unordered_map<int, vector<string>> mp; 
        int index = 0, len = 0;
        for (string& word : words) {
            len += word.size() + 1;
            if (len-1 > maxWidth) {
                ++index;
                len = word.size()+1;
                mp[index].push_back(word);
            } else if (len-1 == maxWidth) {
                len = 0;
                mp[index].push_back(word);
                ++index;
            } else {
                mp[index].push_back(word);
            }
        }

        int lm = mp.size();
        for (int i = 0; i < lm-1; ++i) {
            int ls = mp[i].size();
            int length = 0;
            for (string& s : mp[i]) {
                length += s.size();
            }
            length = maxWidth - length;
            string s = "";
            for (string& m : mp[i]) {
                s += m;
                float f = ls-1 == 0 ? ls : (float)length / (float)(ls-1);
                int space = f;
                if (f > space) ++space;
                --ls;
                length - space < 0 ? s += string(length, ' ') : s += string(space, ' ');
                length -= space;
            }
            if (length > 0) s += string(length, ' ');
            res.push_back(s);
        }
        string tmp = "";
        for (int i = 0; i < mp[lm-1].size(); ++i) {
            if (i != mp[lm-1].size()-1) {
                tmp += mp[lm-1][i] + ' ';
            } else {
                tmp += mp[lm-1][i];
            }
        }
        tmp += string(maxWidth-tmp.size(), ' ');
        res.push_back(tmp);
        return res;
    }
};