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本文小编为大家详细介绍“C++如何实现文本左右对齐”,内容详细,步骤清晰,细节处理妥当,希望这篇“C++如何实现文本左右对齐”文章能帮助大家解决疑惑,下面跟着小编的思路慢慢深入,一起来学习新知识吧。
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 becase 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 "
]
我将这道题翻译为文本的左右对齐是因为这道题像极了word软件里面的文本左右对齐功能,这道题我前前后后折腾了快四个小时终于通过了OJ,完成了之后想着去网上搜搜看有没有更简单的方法,搜了一圈发现都差不多,都挺复杂的,于是乎就按自己的思路来说吧,由于返回的结果是多行的,所以我们在处理的时候也要一行一行的来处理,首先要做的就是确定每一行能放下的单词数,这个不难,就是比较n个单词的长度和加上n - 1个空格的长度跟给定的长度L来比较即可,找到了一行能放下的单词个数,然后计算出这一行存在的空格的个数,是用给定的长度L减去这一行所有单词的长度和。得到了空格的个数之后,就要在每个单词后面插入这些空格,这里有两种情况,比如某一行有两个单词"to" 和 "a",给定长度L为6,如果这行不是最后一行,那么应该输出"to a",如果是最后一行,则应该输出 "to a ",所以这里需要分情况讨论,最后一行的处理方法和其他行之间略有不同。最后一个难点就是,如果一行有三个单词,这时候中间有两个空,如果空格数不是2的倍数,那么左边的空间里要比右边的空间里多加入一个空格,那么我们只需要用总的空格数除以空间个数,能除尽最好,说明能平均分配,除不尽的话就多加个空格放在左边的空间里,以此类推,具体实现过程还是看代码吧:
class Solution { public: vector<string> fullJustify(vector<string> &words, int L) { vector<string> res; int i = 0; while (i < words.size()) { int j = i, len = 0; while (j < words.size() && len + words[j].size() + j - i <= L) { len += words[j++].size(); } string out; int space = L - len; for (int k = i; k < j; ++k) { out += words[k]; if (space > 0) { int tmp; if (j == words.size()) { if (j - k == 1) tmp = space; else tmp = 1; } else { if (j - k - 1 > 0) { if (space % (j - k - 1) == 0) tmp = space / (j - k - 1); else tmp = space / (j - k - 1) + 1; } else tmp = space; } out.append(tmp, " "); space -= tmp; } } res.push_back(out); i = j; } return res; } };
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