LeetCode 68. 文本左右对齐（字符串逻辑题）

1. 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                  "
]

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。

2. 解题

class Solution {	// C++
public:vector<string> fullJustify(vector<string>& words, int maxWidth) {vector<string> ans;string line;int i, width = 0, wc;for(i = 0; i < words.size(); ++i){if(line.empty()){	//为空，直接加入单词line = words[i];width = words[i].size();wc = 1;//单词个数}else{if(width+1+words[i].size() <= maxWidth){	//还能加入line += " "+words[i];width += 1+words[i].size();wc++;}else//超了，放不下i{process(wc,line,maxWidth,width);//处理单词ans.push_back(line);//该行存入答案line = "";width = wc = 0;i--;}}}line += string(maxWidth-width,' ');//最后一行左对齐，后面补空格ans.push_back(line);return ans;}void process(int wc, string& line, int maxWidth, int width){if(wc == 1)//只有一个单词，直接后面补空格{line += string(maxWidth-width,' ');return;}int space = maxWidth - width;//需要的空格数int n = space/(wc-1);//平均插入个数int pos = wc-1;//可以插入的位置个数for(int i = line.size()-1; i >= 0; --i){if(line[i] == ' '){	//找到空格了line.insert(i,n,' ');//插入平均的个数space -= n;//空格数更新pos--;//位置数更新if(pos > 0 && space%pos == 0)//位置还有，且能被整除n = space/pos;//变成整除的（左边空格大于右边条件）}}}
};

0 ms 7 MB

class Solution:# py3def fullJustify(self, words: List[str], maxWidth: int) -> List[str]:ans = []line = ""width = 0wc = 0def process(wc,line,width):if wc==1:line += ' '*(maxWidth-width)return linespace = maxWidth-widthn = space//(wc-1)pos = wc-1line = list(line)size = len(line)for i in range(size-1,-1,-1):if line[i]==' ':line.insert(i, ' '*n)space -= npos -= 1if pos > 0 and space%pos==0:n = space//posline = "".join(line)return linei = 0while i < len(words):if len(line)==0:line = words[i]width = len(words[i])wc = 1else:if width+1+len(words[i]) <= maxWidth:line += " "+words[i]width += 1+len(words[i])wc += 1else:temp = process(wc,line,width)ans.append(temp)line = ""width, wc = 0, 0i -= 1i += 1line += ' '*(maxWidth-width)ans.append(line)return ans