面试经典150题——Day24

文章目录

- 一、题目
- 二、题解

一、题目

68. Text Justification

Given an array of strings 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 does 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 because 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 "

Constraints:

1 <= words.length <= 300

1 <= words[i].length <= 20

words[i] consists of only English letters and symbols.

1 <= maxWidth <= 100

words[i].length <= maxWidth

题目来源： leetcode

二、题解

cpp 复制代码

class Solution {
    // blank 返回长度为 n 的由空格组成的字符串
    string blank(int n) {
        return string(n, ' ');
    }

    // join 返回用 sep 拼接 [left, right) 范围内的 words 组成的字符串
    string join(vector<string> &words, int left, int right, string sep) {
        string s = words[left];
        for (int i = left + 1; i < right; ++i) {
            s += sep + words[i];
        }
        return s;
    }

public:
    vector<string> fullJustify(vector<string> &words, int maxWidth) {
        vector<string> ans;
        int right = 0, n = words.size();
        while (true) {
            int left = right; // 当前行的第一个单词在 words 的位置
            int sumLen = 0; // 统计这一行单词长度之和
            // 循环确定当前行可以放多少单词，注意单词之间应至少有一个空格
            while (right < n && sumLen + words[right].length() + right - left <= maxWidth) {
                sumLen += words[right++].length();
            }

            // 当前行是最后一行：单词左对齐，且单词之间应只有一个空格，在行末填充剩余空格
            if (right == n) {
                string s = join(words, left, n, " ");
                ans.emplace_back(s + blank(maxWidth - s.length()));
                return ans;
            }

            int numWords = right - left;
            int numSpaces = maxWidth - sumLen;

            // 当前行只有一个单词：该单词左对齐，在行末填充剩余空格
            if (numWords == 1) {
                ans.emplace_back(words[left] + blank(numSpaces));
                continue;
            }

            // 当前行不只一个单词
            int avgSpaces = numSpaces / (numWords - 1);
            int extraSpaces = numSpaces % (numWords - 1);
            string s1 = join(words, left, left + extraSpaces + 1, blank(avgSpaces + 1)); // 拼接额外加一个空格的单词
            string s2 = join(words, left + extraSpaces + 1, right, blank(avgSpaces)); // 拼接其余单词
            ans.emplace_back(s1 + blank(avgSpaces) + s2);
        }
    }
};