LeetCode: Text Justification

最新推荐文章于 2020-06-10 05:27:24 发布

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本文介绍了一种算法，用于将一组单词按照指定长度进行格式化处理，确保每行文字达到完全左对齐和右对齐的效果。文章详细解释了如何通过贪婪方式填充单词，并在必要时插入额外空格来实现每行字符数一致的目标。

Given an array of words and a length L, format the text such that each line has exactly L 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 exactlyL 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.

For example,
words: ["This", "is", "an", "example", "of", "text", "justification."]
L: 16.

Return the formatted lines as:

[
   "This    is    an",
   "example  of text",
   "justification.  "
]

Note: Each word is guaranteed not to exceed L in length..

class Solution {
public:
    vector<string> fullJustify(vector<string> &words, int L) {
        int len = 0, count = 0;
        vector<string> result;
        if(words[0] == "")
        {
            string key(L, ' ');
            result.push_back(key);
            return result;
        }
        vector<string> temp;
        string cur;
        for(int i = 0; i < words.size(); i++)
        {
            if(len + words[i].size() + count <= L)
            {
                temp.push_back(words[i]);
                len += words[i].size();
                count++;
                if(i == words.size()-1)
                {
                    int remain = L - (len + count-1);
                    for(int i = 0; i < temp.size()-1; i++)
                    {
                        cur += temp[i];
                        cur += " ";
                    }
                    cur += temp[temp.size()-1];
                    string key(remain, ' ');
                    cur += key;
                    result.push_back(cur);
                    cur = "";
                    count = 0;
                    len = 0;
                    temp.clear();
                }
            }
            else
            {
                int remain = L - (len + count-1);
                if(count == 1)
                {
                    cur += temp[0];
                    string key(remain, ' ');
                    cur += key;
                }
                else
                {
                    int each = remain / (count-1);
                    remain = remain % (count-1);
                    string key(each+1, ' ');
                    for(int i = 0; i < temp.size()-1; i++)
                    {
                        cur += temp[i];
                        cur += key;
                        if(i < remain)
                            cur += " ";
                    }
                    cur += temp[temp.size()-1];
                }
                result.push_back(cur);
                cur = "";
                count = 0;
                len = 0;
                temp.clear();
                i--;
            }
            
        }
        return result;
    }
};