Leetcode 68 Text Justification

最新推荐文章于 2020-09-08 19:13:08 发布

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最新推荐文章于 2020-09-08 19:13:08 发布

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本文链接：https://blog.youkuaiyun.com/Neo233/article/details/85220665

本文介绍了一个文本格式化问题，目标是在给定的最大宽度下，使每行文本包含尽可能多的单词，并确保每行都有确切的最大宽度字符数。讨论了如何在单词间均匀分布额外的空格，以及如何处理最后一行的左对齐要求。

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Given an array of words and a width maxWidth, format the text such that each line has exactly maxWidth characters and is fully (left and right) justified.

You should pack your words in a greedy approach; that is, pack as many words as you can in each line. Pad extra spaces ' ' when necessary so that each line has exactly maxWidth characters.

Extra spaces between words should be distributed as evenly as possible. If the number of spaces on a line do not divide evenly between words, the empty slots on the left will be assigned more spaces than the slots on the right.

For the last line of text, it should be left justified and no extra space is inserted between words.

Note:

A word is defined as a character sequence consisting of non-space characters only.
Each word's length is guaranteed to be greater than 0 and not exceed maxWidth.
The input array words contains at least one word.

Example 1:

Input:
words = ["This", "is", "an", "example", "of", "text", "justification."]
maxWidth = 16
Output:
[
   "This    is    an",
   "example  of text",
   "justification.  "
]

Example 2:

Input:
words = ["What","must","be","acknowledgment","shall","be"]
maxWidth = 16
Output:
[
  "What   must   be",
  "acknowledgment  ",
  "shall be        "
]
Explanation: Note that the last line is "shall be    " instead of "shall     be",
             because the last line must be left-justified instead of fully-justified.
             Note that the second line is also left-justified becase it contains only one word.

Example 3:

Input:
words = ["Science","is","what","we","understand","well","enough","to","explain",
         "to","a","computer.","Art","is","everything","else","we","do"]
maxWidth = 20
Output:
[
  "Science  is  what we",
  "understand      well",
  "enough to explain to",
  "a  computer.  Art is",
  "everything  else  we",
  "do                  "
]

这个题根据题意来模拟就行

class Solution {
public:
    string space(int x){
        string res;
        while(x--){
            res += ' ';
        }
        return res;
    }
    vector<string> fullJustify(vector<string>& words, int maxWidth) {
        vector<string> res;
        for(int i = 0 ;i < words.size();){
            int j = i + 1,s = words[i].size(),rs = words[i].size();
            while(j < words.size() && s + 1 + words[j].size() <= maxWidth){
                s += words[j].size() + 1;
                rs += words[j].size();
                j++;
            }
            rs = maxWidth - rs;//求一下现在需要填补空格的个数
            string line = words[i];
            if(j == words.size()){//如果现在是最后一行了
                for(int k = i + 1; k < j;k++){
                    line += ' ' + words[k];//放一个空格和单词
                }
                line += space(maxWidth - line.size());//需要补满空格
            }else if(j - i ==1){//里面只有一个单词
                line += space(maxWidth - line.size());
            }else{
                int base = rs / (j - i - 1);//算一下每个单词之间放多少个空格 
                int rem = rs % (j - i - 1);
                i++;
                for(int k = 0;i < j;i++,k++){
                    line += space(base +(k < rem)) + words[i];//加上空格和第i个单词
                }
            }
            i = j;//处理结束了
            res.push_back(line);//将这一行加到结果中
        }
        return res;
    }
};