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 " ]
class Solution {
public:
vector<string> fullJustify(vector<string>& words, int maxWidth) {
int n = words.size();
vector<string> result;
for (int i = 0; i < n; i++)
{
int begin = i;
int wordLen = words[i].size();
while (i + 1 < n && words[i + 1].size() + wordLen + 1 <= maxWidth)
{
wordLen += words[++i].size() + 1;
}
int numberofWords = i - begin + 1;
int space = 1;
int extraSpace = 0;
if (numberofWords > 1 && i < n - 1)
{
int remaining = maxWidth - wordLen;
space = remaining / (numberofWords - 1) + 1;
extraSpace = remaining % (numberofWords - 1);
}
string line = words[begin];
for (int j = 1; j < numberofWords; j++)
{
line.append(space, ' ');
if (j <= extraSpace)
{
line.push_back(' ');
}
line += words[begin + j];
}
if (line.size() < maxWidth)
{
line.append(maxWidth - line.size(), ' ');
}
result.emplace_back(line);
}
return result;
}
};