0068.文本左右对齐.cpp

/*
 * @lc app=leetcode.cn id=68 lang=cpp
 *
 * [68] 文本左右对齐
 *
 * https://leetcode-cn.com/problems/text-justification/description/
 *
 * algorithms
 * Hard (40.57%)
 * Likes:    40
 * Dislikes: 0
 * Total Accepted:    4.7K
 * Total Submissions: 11.3K
 * Testcase Example:  '["This", "is", "an", "example", "of", "text", "justification."]\n16'
 *
 * 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                  "
 * ]
 * 
 * 
 */

// @lc code=start
#include <string>
#include <vector>
using namespace std;

class Solution {
public:
    vector<string> fullJustify(vector<string>& words, int maxWidth)
    {
        vector<string> ans;
        vector<string> tmp;
        string row;
        int word_cnt = 0;
        int left_len = maxWidth;
        int space_size = 0;
        int len;
        for (int i = 0; i < words.size(); ++i) {
            len = words[i].size();
            if (len <= left_len) {
                tmp.push_back(words[i]);
                left_len -= (len + 1);
                ++word_cnt;
            } else {
                --i;
                ++left_len;
                if (word_cnt == 1) {
                    row += tmp[0] + string(left_len, ' ');
                } else {
                    space_size = left_len / (word_cnt - 1);
                    for (int j = 0; j < tmp.size() - 1; ++j) {
                        tmp[j] += string(space_size + 1, ' ');
                    }
                    for (int j = 0; j < left_len % (word_cnt - 1); ++j) {
                        tmp[j] += ' ';
                    }
                    for (auto s : tmp) {
                        row += s;
                    }
                }
                tmp.clear();
                ans.push_back(row);
                row = "";
                left_len = maxWidth;
                word_cnt = 0;
            }
            if (i == words.size() - 1 && tmp.size() != 0) {
                for (int j = 0; j < tmp.size() - 1; ++j)
                    row += tmp[j] + " ";
                row += tmp[tmp.size() - 1] + string(left_len + 1, ' ');
                ans.push_back(row);
            }
        }
        return ans;
    }
};

// @lc code=end