LeetCode: 557Reverse Words in a String III(easy)

玲珑子_a

于 2019-04-17 14:25:13 发布

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分类专栏： LeetCode学习笔记

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给定一个字符串，你需要反转每个单词内的字符顺序，同时保持空格和初始单词顺序不变。例如：输入'Let's take LeetCode contest'，输出'stleL ekat edoCteL ecnats'。题目中指出字符串中单词由单个空格分隔，且无额外空格。注意到原代码运行时间为772ms，而其他人的实现为26ms，说明使用栈可能会导致效率较低。

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题目：

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.

代码：

class Solution {
public:
    string reverseWords(string s) {
        stack<char> tem;
        string result;
        string s1 = s + ' ';
        for (auto c : s1){
            if ( c != ' ')
                tem.push(c);
            else {
                while(!tem.empty()){
                    result = result + tem.top();
                    tem.pop();
                }
                result = result + ' ';
            }
        }
        result = result.substr(0, result.length()-1);
        return result;
    }
};

运行时间：772ms

别人的：

class Solution {
public:
    
    string reverseWords(string s) {
        string result(s.length(), 32);
        int start = 0;
        int end = s.find(32);
        while (end != -1) {
            for (int i = start; i < end; ++i)
                result[end - (i - start) - 1] = s[i];
            start = end + 1;
            end = s.find(32, start);
        }
        end = s.length();
        for (int i = start; i < end; ++i)
            result[end - (i - start) - 1] = s[i];
        return result;
    }
};

运行时间：26mm

栈比较耗时间？