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.
题目的要求很简单，要求翻转每个单词，但是保持空格以及单词的相对顺序不变，最开始我想的是用一个栈，遇见空格把栈中的单词弹出，最后也确实AC了，代码如下：
1 class Solution { 2 public: 3 string reverseWords(string s) { 4 stack<char> sta; 5 //s += ‘ ‘; 6 string res = ""; 7 for (int i = 0; i <= s.length(); i++) 8 { 9 if (s[i] != ‘ ‘ && i != (s.length())) 10 //if (s[i] != ‘ ‘) 11 { 12 sta.push(s[i]); 13 //cout<<sta.top()<<endl; 14 } 15 else 16 { 17 while (!sta.empty()) 18 { 19 res += sta.top(); 20 //std::cout<<sta.top(); 21 sta.pop(); 22 } 23 if (s[i] == ‘ ‘) 24 res += ‘ ‘; 25 } 26 } 27 cout<<endl; 28 return res; 29 } 30 };
当然也可以使用reverse函数，代码如下：
class Solution { public: string reverseWords(string s) { int front = 0; for (int i = 0; i <= s.length(); i++) { if (s[i] == ‘ ‘ || i == s.length()) { reverse(&s[front], &s[i]); front = i + 1; } } return s; } };
或者是使用两个指针，分别指向字符串的 头和尾部：
1 class Solution { 2 public: 3 string reverseWords(string s) { 4 int start = 0, end = 0, n = s.size(); 5 while (start < n && end < n) { 6 while (end < n && s[end] != ‘ ‘) ++end; 7 for (int i = start, j = end - 1; i < j; ++i, --j) { 8 swap(s[i], s[j]); 9 } 10 start = ++end; 11 } 12 return s; 13 } 14 };
