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题目
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 " ]
解法
class Solution {
public List<String> fullJustify(String[] words, int maxWidth) {
List<String> rList = new ArrayList<>();
List<String> list = new ArrayList<>();
int count = 0;
count = words[0].length();
list.add(words[0]);
for(int i = 1; i < words.length; i++){
if(count + words[i].length() + 1 <= maxWidth){ //判断剩余位置是否足够放一个空格以及下一个单词
count = count + words[i].length() + 1;
list.add(" " + words[i]);
continue;
}
//位置不足，生成一行字符串
StringBuilder b = new StringBuilder();
if(list.size() == 1){ //只有一个单词，在后方补足空格
b.append(list.get(0));
while(b.length() < maxWidth){
b.append(" ");
}
}else if(list.size() == 2){//只有两个空格，在中部补足空格
b.append(list.get(0));
while(b.length() + list.get(1).length() < maxWidth){
b.append(" ");
}
b.append(list.get(1));
}else{
int m = (maxWidth - count)/(list.size()-1);//每个单词间要补多少个空格
int n = (maxWidth - count)%(list.size()-1);//前多少个单词间要额外补一个空格
for(int j = 0; j < list.size() - 1; j++){
b.append(list.get(j));
for(int z = 0; z < m; z++){
b.append(" ");
}
if(j<n){
b.append(" ");
}
}
b.append(list.get(list.size()-1));
}
rList.add(b.toString());
list.clear();
count = words[i].length();
list.add(words[i]);
}
//生成最后一行
StringBuilder b = new StringBuilder();
for(int j = 0; j < list.size(); j++){
b.append(list.get(j));
}
while(b.length()<maxWidth){
b.append(" ");
}
rList.add(b.toString());
return rList;
}
}