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LeetCode 算法题 68. 文本左右对齐

分类：algorithm| 发布时间：2017-04-05 11:12:00

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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                  "
]

解法

class Solution {
public:
    vector<string> fullJustify(vector<string>& words, int maxWidth) {
        vector<string> ret;
        int currentLen = 0, begin = 0, i = 0;
        for (; i < words.size(); ++i) {
            if (currentLen + words[i].size() + (i - begin) > maxWidth) {
                packToLine(words, begin, i, currentLen, maxWidth, ret);
                begin = i;
                currentLen = words[i].size();
            } else {
                currentLen += words[i].size();
            }
        }

        lastPack(words, begin, words.size(), currentLen, maxWidth, ret);
        return ret;
    }

    void packToLine(vector<string> &words, int begin, int end, int currentLen, int maxWidth, vector<string> &ret) {
        const int space = maxWidth - currentLen;
        if (end == begin + 1) {
            string s = words[begin];
            s.append(space, ' ');
            ret.push_back(s);
            return;
        }

        int pad = space / (end - begin - 1);
        int extra = space % (end - begin - 1);
        string s;
        for (int i = begin; i < end - 1; ++i) {
            s += words[i];
            s.append(pad, ' ');
            if (extra) {
                s.push_back(' ');
                --extra;
            }
        }

        s += words[end - 1];
        ret.push_back(s);
    }

    void lastPack(vector<string> &words, int begin, int end, int currentLen, int maxWidth, vector<string> &ret) {
        int space = maxWidth - currentLen;
        string s;
        for (int i = begin; i < end; ++i) {
            s += words[i];
            if (space) {
                s.push_back(' ');
                --space;
            }
        }

        if (space)
            s.append(space, ' ');

        ret.push_back(s);
    }
};