text-justification

最新推荐文章于 2018-11-16 14:32:58 发布

llwjason5555

最新推荐文章于 2018-11-16 14:32:58 发布

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分类专栏： leetcode

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本文介绍了一个用于文本格式化的算法，该算法能够确保每行文本恰好包含指定数量的字符，并实现左右两端对齐的效果。通过贪婪策略填充尽可能多的单词到一行中，并在必要时添加额外的空格来调整每行的字符数。

题目：

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 exactly L 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.
click to show corner cases.
Corner Cases:

A line other than the last line might contain only one word. What should you do in this case?
In this case, that line should be left-justified.

程序：

class Solution {
public:
    vector<string> fullJustify(vector<string> &words, int L) {
        int len = words.size();
        if (len == 0||L<=0)
            return words;

        vector<string> res;
        string line;
        int i = 0;
        while (i < len)
        {
            int n = 0;//文字长度
            int j = 0;//单词个数
            while (n+j-1 <= L&&i<len)
            {
                n = n + words[i].size();
                i++;
                j++;

            }

            if (n+j-1>L)
            {
                i--;
                j--;
                n = n - words[i].size();
            }


            if (j == 1 && i < len)
            {
                int length = L - n;//只有一个单词的特殊情况
                string str(length, ' ');
                line = words[i - 1] + str;
                res.push_back(line);
                line.clear();
                continue;
            }
            if (i == len)//行尾
            {
                for (int k = j - 1; k >= 0; k--)
                {
                    if (k)
                        line = line + words[i - (k + 1)] + " ";
                    else
                        line = line + words[i - (k + 1)];
                }
                if(line.size()<L)
                {
                    string space(L-line.size(),' ');
                    line=line+space;
                }
                res.push_back(line);
                line.clear();
                break;
            }
            int length = L - n;//空格长度
            int ave;//平均空格长度
            int left;//多余空格长度

            if (length)
            {
                 ave = length / (j - 1);//平均空格
                 left = length % (j - 1);//多余出来的空格靠左的每个加一
            }
            string str(ave, ' ');
            for (int k = j-1; k>=0; k--)
            {
                if (k != 0)
                    line = line + words[i-(k+1)] + str;
                else
                    line = line + words[i-(k+1)];
                if (left)
                {
                    line = line + " ";
                    left--;
                }
            }
            res.push_back(line);
            line.clear();



        }
        return res;
    }
};