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 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 <= 3001 <= words[i].length <= 20words[i]由小写英文字母和符号组成1 <= maxWidth <= 100words[i].length <= maxWidth
class Solution {
public List<String> fullJustify(String[] words, int maxWidth) {
/**
* special:
* lastLine or only one word in a line
*/
int max, len = words.length, start, end = 0, spaceNum, wordNum;
List<String> res = new ArrayList<>();
while (true) {
start = end;
max = 0;
while (end < len && max + words[end].length()+(end-start) <= maxWidth){
max += words[end++].length();
}
/**
* lastLine
*/
if (end == len){
StringBuffer sb = new StringBuffer(words[start]);
for (int i = start+1; i < end; i++) {
sb.append(' ');
sb.append(words[i]);
}
int length = sb.length();
for (int i = 0; i < maxWidth-length; i++) {
sb.append(' ');
}
res.add(sb.toString());
return res;
}
wordNum = end - start;
spaceNum = maxWidth - max;
/**
* only one word in a line
*/
if (wordNum == 1) {
StringBuffer sb = new StringBuffer(words[start]);
for (int i = 0; i < spaceNum; i++) {
sb.append(' ');
}
res.add(sb.toString());
continue;
}
/**
* common situation
*/
StringBuffer sb = new StringBuffer(words[start]);
for (int i = start+1; i < end; i++) {
if (spaceNum % (wordNum-1) == 0) {
for (int j = 0; j < spaceNum/ (wordNum-1); j++) {
sb.append(' ');
}
}else {
spaceNum -= 1;
sb.append(' ');
for (int j = 0; j < spaceNum/ (wordNum-1); j++) {
sb.append(' ');
}
}
sb.append(words[i]);
}
res.add(sb.toString());
}
}
}
