[Leet Code] Reverse String(Java)

初试Java写Leet

public class Solution {
    public String reverseString(String s) {
        char[] stringArr= s.toCharArray();
        char newS;
        int last = stringArr.length-1;
        for(int i=0;i<stringArr.length/2;i++){
            newS = stringArr[i];
            stringArr[i] = stringArr[last-i];
            stringArr[last-i] = newS;
        }
        
        return new String(stringArr);
    }
}


<div class="content" itemprop="text"><p>public class Solution {</p><pre class="markdown-highlight"><code class="hljs java"><span class="hljs-function"><span class="hljs-keyword">public</span> String <span class="hljs-title">reverseString</span><span class="hljs-params">(String s)</span> </span>{
    <span class="hljs-keyword">if</span>(s.length() < <span class="hljs-number">2</span>) {
        <span class="hljs-keyword">return</span> s;
    }
    
    StringBuilder result = <span class="hljs-keyword">new</span> StringBuilder();
    
    <span class="hljs-keyword">for</span>(<span class="hljs-keyword">int</span> i = s.length()-<span class="hljs-number">1</span>; i >= <span class="hljs-number">0</span>; i--) {
        result.append(s.charAt(i));
    }
    
    <span class="hljs-keyword">return</span> result.toString();
}
</code>

}


 

粘上代码,原本答案是O(n)但时间复杂度超过限定,大脑短路,遂抄袭论坛


在来一段Python大法的吊炸天代码

class Solution(object):
    def reverseString(self, s):
        n = len(s)
        s = list(s)
        
         # return self.reverseStringRecurse(s, 0, len(s)-1)
         # return self.reverseString2(s)
         # return self.reverseString3(s)
        
        for i in range(n/2):
            s[i], s[~i] = s[~i], s[i]
        
        return "".join(s)
            
    def reverseString2(self, s):
        return s[::-1] 
    
    def reverseString3(self, s):
        return "".join(reversed(list(s))) 
        
    def reverseStringRecurse(self, s, lo=0,hi=None):
        if hi<=lo:
            return "".join(s)
        
        s[lo],s[hi] = s[hi], s[lo]
        return self.reverseStringRecurse(s, lo+1, hi-1)


看了论坛的代码真是无地自容啊!!!

2016/8/23

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