Question

以下算法的上限是什么，它会反转给定句子的每个单词：

for i = 1 to n
     if(space is found)
         reverse(word)

例如，句子=＆＃34;运行时分析＆＃34;

=＆gt;输出将是＆＃34; nuR emiT sisylanA＆＃34;

将是O（n ^ 2）？还是O（n）？假设 reverse（word） 运行一个字长循环。

Answer 1

答案是O（n），因为即使你必须反转过去的字符串，迭代次数将是O（2n）但是2n是可折旧的，所以O（n）是答案，因为当n很大时，因子2是可折旧的。

Answer 2

这是mi证明

#include <bits/stdc++.h>
using namespace std;

int iterations; // Number of iteration that the code will run
string sentence; //The sentece that you want to reverse

void reverse (int start, int end) 
{
    for (int i = start, k = 0; k <= ((end-start)/2); i++, k++) {
        //swap( sentence[i], sentence[end-i] );
        /* This is a swap */
        char keep = sentence[i];
        sentence[i] = sentence[(end-k)];
        sentence[(end-k)] = keep;
        iterations++;
    }

}
 //4 - 7 time 7 - 4 = 3/2 = 1 
int main() {

    sentence = "Run Time Analysis";
    string origin = sentence;
    int len = sentence.length(), start = 0, end = 0;

    iterations = 0; //Starts from 0

    for (int i = 0; i < len; i++) {
        if (sentence[i] == ' '  || i == (len-1)) {
            i = (i==len-1) ? (i+1) : i;
            end = i-1;
            reverse(start, end);    
            start = i+1;
        }
        iterations++;
    }
    cout  << "Orginal sentence: " << origin << "\nResult: " << sentence << "\nLength of the sentence: " << len << "\nNumber of iterations: " << iterations << "\n";
    return 0;
}

执行此算法的结果是 O（n） http://ideone.com/1I4QCY 。如果这不能说服你那么我就不知道了。

<强> RESULT

Orginal sentence: Run Time Analysis, 
Result: nuR emiT sisylanA, 
Length of the sentence: 17, 
Number of iterations: 25

big-O中的运行时间分析

2 个答案: