[LeetCode] Reverse Words in a String III 翻轉字串中的單詞之三
阿新 • • 發佈:2018-12-27
Given a string, you need to reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order.
Example 1:
Input: "Let's take LeetCode contest" Output: "s'teL ekat edoCteeL tsetnoc"
Note: In the string, each word is separated by single space and there will not be any extra space in the string.
這道題讓我們翻轉字串中的每個單詞,感覺整體難度要比之前兩道Reverse Words in a String II和Reverse Words in a String要小一些,由於題目中說明了沒有多餘空格,使得難度進一步的降低了。首先我們來看使用字元流處理類stringstream來做的方法,相當簡單,就是按順序讀入每個單詞進行翻轉即可,參見程式碼如下:
解法一:
class Solution { public: string reverseWords(string s) { string res = "", t = ""; istringstreamis(s); while (is >> t) { reverse(t.begin(), t.end()); res += t + " "; } res.pop_back(); return res; } };
下面我們來看不使用字元流處理類,也不使用STL內建的reverse函式的方法,那麼就是用兩個指標,分別指向每個單詞的開頭和結尾位置,確定了單詞的首尾位置後,再用兩個指標對單詞進行首尾交換即可,有點像驗證迴文字串的方法,參見程式碼如下:
解法二:
class Solution { public: string reverseWords(string s) { int start = 0, end = 0, n = s.size(); while (start < n && end < n) { while (end < n && s[end] != ' ') ++end; for (int i = start, j = end - 1; i < j; ++i, --j) { swap(s[i], s[j]); } start = ++end; } return s; } };
類似題目:
參考資料: