1 public String reverse(String src) { 2 char[] value = src.toCharArray(); 3 int count = value.length; 4 int n = count - 1; 5 for (int j = (n - 1) >> 1; j >= 0; j--) { 6 char tmp = value[j]; 7 char tmp2 = value[n - j]; 8 value[j] = tmp2; 9 value[n - j] = tmp; 10 } 11 return new String(value); 12 }
替换原则：index k 的值和 n-k
的值进行交换。（始终记住程序员的n、k都是字符串的实际位置。）
乘除的最基本实现还是来源于移位操作。
Java字符串反转有2个已经实现的类函数： StringBuffer、StringBuilder类的函数reverse() 。
其中源码的实现如下，核心的思想是替换原则：
View Code
1 /** 2 * Causes this character sequence to be replaced by the reverse of 3 * the sequence. If there are any surrogate pairs included in the 4 * sequence, these are treated as single characters for the 5 * reverse operation. Thus, the order of the high-low surrogates 6 * is never reversed. 7 * 8 * Let <i>n</i> be the character length of this character sequence 9 * (not the length in <code>char</code> values) just prior to 10 * execution of the <code>reverse</code> method. Then the 11 * character at index <i>k</i> in the new character sequence is 12 * equal to the character at index <i>n-k-1</i> in the old 13 * character sequence. 14 * 15 * <p>Note that the reverse operation may result in producing 16 * surrogate pairs that were unpaired low-surrogates and 17 * high-surrogates before the operation. For example, reversing 18 * "\uDC00\uD800" produces "\uD800\uDC00" which is 19 * a valid surrogate pair. 20 * 21 * @return a reference to this object. 22 */ 23 public AbstractStringBuilder reverse() { 24 boolean hasSurrogate = false; 25 int n = count - 1; 26 for (int j = (n-1) >> 1; j >= 0; --j) { 27 char temp = value[j]; 28 char temp2 = value[n - j]; 29 if (!hasSurrogate) { 30 hasSurrogate = (temp >= Character.MIN_SURROGATE && temp <= Character.MAX_SURROGATE) 31 || (temp2 >= Character.MIN_SURROGATE && temp2 <= Character.MAX_SURROGATE); 32 } 33 value[j] = temp2; 34 value[n - j] = temp; 35 } 36 if (hasSurrogate) { 37 // Reverse back all valid surrogate pairs 38 for (int i = 0; i < count - 1; i++) { 39 char c2 = value[i]; 40 if (Character.isLowSurrogate(c2)) { 41 char c1 = value[i + 1]; 42 if (Character.isHighSurrogate(c1)) { 43 value[i++] = c1; 44 value[i] = c2; 45 } 46 } 47 } 48 } 49 return this; 50 }
其中的hasSurrogate与字符编码有关系，这个东东暂时没有研究。