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本文小編為大家詳細(xì)介紹“C++如何實現(xiàn)文本左右對齊”,內(nèi)容詳細(xì),步驟清晰,細(xì)節(jié)處理妥當(dāng),希望這篇“C++如何實現(xiàn)文本左右對齊”文章能幫助大家解決疑惑,下面跟著小編的思路慢慢深入,一起來學(xué)習(xí)新知識吧。
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 "
]
我將這道題翻譯為文本的左右對齊是因為這道題像極了word軟件里面的文本左右對齊功能,這道題我前前后后折騰了快四個小時終于通過了OJ,完成了之后想著去網(wǎng)上搜搜看有沒有更簡單的方法,搜了一圈發(fā)現(xiàn)都差不多,都挺復(fù)雜的,于是乎就按自己的思路來說吧,由于返回的結(jié)果是多行的,所以我們在處理的時候也要一行一行的來處理,首先要做的就是確定每一行能放下的單詞數(shù),這個不難,就是比較n個單詞的長度和加上n - 1個空格的長度跟給定的長度L來比較即可,找到了一行能放下的單詞個數(shù),然后計算出這一行存在的空格的個數(shù),是用給定的長度L減去這一行所有單詞的長度和。得到了空格的個數(shù)之后,就要在每個單詞后面插入這些空格,這里有兩種情況,比如某一行有兩個單詞"to" 和 "a",給定長度L為6,如果這行不是最后一行,那么應(yīng)該輸出"to a",如果是最后一行,則應(yīng)該輸出 "to a ",所以這里需要分情況討論,最后一行的處理方法和其他行之間略有不同。最后一個難點就是,如果一行有三個單詞,這時候中間有兩個空,如果空格數(shù)不是2的倍數(shù),那么左邊的空間里要比右邊的空間里多加入一個空格,那么我們只需要用總的空格數(shù)除以空間個數(shù),能除盡最好,說明能平均分配,除不盡的話就多加個空格放在左邊的空間里,以此類推,具體實現(xiàn)過程還是看代碼吧:
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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