该算法的时间复杂度：是O（n ^ 2）还是O（n）

Question

I am trying to solve "Reverse Words in a String III" of Leetcode. I come up with a solution but I think that my time complexity is O(n^2). My code is able to pass all the test cases. My code is as follows:

class Solution {
    public String reverseWords(String s) {
        if(s.isEmpty())
            return "";

        StringBuilder result = new StringBuilder();        
        String[] str = s.split("\\s+");
        for(String s1:str){
            char[] c1  = reverseChar(s1);
            result.append(c1).append(" ");
            }
        return result.toString().trim();
    }

    public char[] reverseChar(String s){
        char[] c = s.toCharArray();
        int i = 0;
        int j = c.length-1;

        while(i<j){
                char temp = c[i];
                c[i] = c[j];
                c[j] = temp;
                j--;
                i++;
            }
        return c;
    }

}

Answer 1

O(n) ... and actually slightly less when the String length is short due to start up / tear down costs of the JVM.

If you're interested in knowing precisely, I recommend researching the Java Micro-benchmarking Harness (JMH) (re: https://www.oracle.com/technetwork/articles/java/architect-benchmarking-2266277.html ).

I do not recommend the following poor-man's benchmark; however, it answers the question all the same.

    public static void main (String[] args) {
        Random r = new Random();
        char[] chars = new char[100_000]; // change this value to other benchmarks
                for (char c : chars) {
            c = (char) (r.nextInt(90) + 32);
        }

        Solution s = new Solution();
        long x1  = System.currentTimeMillis();
        s.reverseWords(Arrays.toString(chars));
        long x2 = System.currentTimeMillis();

        System.out.println("duration = " + (x2 - x1));
    }

Answer 2

I think the complexity is O(n/2). The code can be abstract as

for(int i =0 ,i< str.length, i++){
    for(int j = 0,j < str[i].length/2, j++)
    {
       ...
    }
}

Say str is ['abc','abc','abc','abc'] ,the complexity is 4*(3/2) = (4*3)/2 .And 4*3 == n ,So it is O(n/2)