【Leetcode】68. Text Justification-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_29600137/article/details/88892145

本文介绍了一种文本排版算法，该算法将一串字符串按指定宽度进行格式化，确保每行字符数量固定，并且单词间的空格分布尽可能均匀。特别地，最后一行左对齐，不添加额外空格。

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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                  "
]

题目大意：

将给出的一串字符串，按照一行maxWidth个输出，每个单词之间需要给出空格。每个单词空格的个数要保持最大平均。

解题思路：

主要是找出每一行能够放入多少个单词，计算出空格的个数。遍历每一行包括的单词每次加一个空格，直达空格数耗尽，注意每一行最后一个单词之后不加空，但是最后一行例外。

class Solution {
public:
    vector<string> fullJustify(vector<string>& words, int maxWidth) {
        vector<string> ans;
        if(maxWidth == 0) return {""};
        int i=0, j=0;
        while(j!=words.size()){
            int len = -1;
            while(j < words.size() && len + words[j].size() + 1<=maxWidth){
                len += words[j++].size() + 1;
            }
            int space = maxWidth - len + j - i -1;
            int k = i;
            while(space){
                words[k++]+=" ";
                space--;
                if(j != words.size() && (k==j-1||k==j)) k=i;
                if(j == words.size() && k==j) k=j-1;
            }
            string line = "";
            for(int l=i;l<j;l++){
                line += words[l];
            }
            ans.push_back(line);
            i = j;
        }
        return ans;
    }
};