整形转换为字符串，写了五种算法，并进行了性能分析：

1. 算法： 直接先求得长度，然后从左向右求得各位的大小

 private static class Int2String1 implements Int2String {public String int2String(int intValue) {if (intValue == 0) {return "0";}int length = 0;boolean sign = false;if (intValue < 0) {intValue = -intValue;sign = true;length++;}int k = 1;while (true) {if (intValue >= k) {length++;k = k * 10;} else {break;}}char[] holder = new char[length];int start = 0;if (sign) {holder[0] = '-';start = 1;}for (; start < length; ++start) {int count = (length - start - 1);int pow10 = 1;for (int i = 0; i < count; i++) {pow10 = pow10 * 10;}holder[start] = (char) ((intValue / Math.pow(10, count)) + '0');intValue = intValue % pow10;}return new String(holder);}}

打印时间：

1the time is 12122, strValue = -100121312

2. 算法： 直接从右向左， 或者说低位到高位进行逐个插入。最后插入符号位

  private static class Int2String2 implements Int2String {public String int2String(int intValue) {char[] str = new char[0];boolean sign = false;if (intValue == 0) {return "0";}if (intValue < 0) {sign = true;intValue = -intValue;}while (true) {int leftValue = intValue % 10;char additionalChar = (char) (leftValue + '0');char[] old = str;str = new char[str.length + 1];str[0] = additionalChar;for (int i = 1; i < str.length; ++i) {str[i] = old[i - 1];}intValue = intValue / 10;if (intValue == 0) {break;}}if (sign) {char[] old = str;str = new char[str.length + 1];str[0] = '-';for (int i = 1; i < str.length; ++i) {str[i] = old[i - 1];}}return new String(str);}}

打印时间：

1the time is 3426, strValue = -100121312

3.算法：先斤算长度，计算长度时与10的倍数比较。再从右向左进行运算各位的值。

   private static class Int2String3 implements Int2String {public String int2String(int intValue) {if (intValue == 0) {return "0";}int length = 0;boolean sign = false;if (intValue < 0) {intValue = -intValue;sign = true;length++;}int k = 1;while (true) {if (intValue >= k) {length++;k = k * 10;} else {break;}}char[] holder = new char[length];int start = 0;if (sign) {holder[0] = '-';start = 1;}for (int i = length - 1; i >= start; --i) {holder[i] = (char) ((intValue % 10) + '0');intValue = intValue / 10;}return new String(holder);}}

打印时间：

1the time is 834, strValue = -100121312

4. 先斤算长度，再从右向左进行运算各位的值。计算长度采用除法求模。

 private static class Int2String4 implements Int2String {public String int2String(int intValue) {int length = 0;boolean sign = false;if (intValue < 0) {sign = true;intValue = -intValue;}int k = intValue;while (k >= 0) {length++;k = k / 10;if (k == 0) {break;}}char[] charArray;if (sign) {length = length + 1;charArray = new char[length];charArray[0] = '-';} else {charArray = new char[length];}int m = intValue;int index = length - 1;while (true) {int leftValue = m % 10;charArray[index] = (char) (leftValue + '0');index--;m = m / 10;if (m == 0) {break;}}return new String(charArray);}}

打印时间：

1the time is 1102, strValue = -100121312

5. 不再计算长度，而是采用一个buffer。最后构造字符串直接使用此buffer构造。由于10进制的位数一定是小于8进制的位数。32位int转换到8进制就是11位，推断10进制的最大位数也是11位。

    private static class Int2String5 implements Int2String {public String int2String(int intValue) {int position = 10;char[] buf = new char[11];if (intValue < 0) {intValue = -intValue;buf[0] = '-';}while (true) {int leftValue = intValue % 10;buf[position] = (char) (leftValue + '0');intValue = intValue / 10;if (intValue == 0) {break;} else {position--;}}return new String(buf, position, buf.length - position);}}

打印时间：

1the time is 742, strValue = 100121312

6. 采用系统自带的算法

 private static class Int2String6 implements Int2String {public String int2String(int intValue) {return String.valueOf(intValue);}}

打印时间：

1the time is 764, strValue = -100121312

由此可见java SDK 提供的算法并不是最优算法。采用buffer避免计算长度的算法速度更快。可以在系统内存要求不高的时候直接提供一个公用的11位的buffer去做转换可能比java系统自带的算法更快。

于是得到了以下算法

   public static char[] buf = new char[11];private static class Int2String7 implements Int2String {public String int2String(int intValue) {int position = 10;if (intValue < 0) {intValue = -intValue;buf[0] = '-';}while (true) {int leftValue = intValue % 10;buf[position] = (char) (leftValue + '0');intValue = intValue / 10;if (intValue == 0) {break;} else {position--;}}return new String(buf, position, buf.length - position);}}

打印：

1the time is 636, strValue = 100121312

可是此算法有个很大的问题，就是在多线程环境下的同步问题，于是解决同步问题，于是有了算法7.

7. 在6的基础上解决同步问题：

    public static char[] buf = new char[11];private static class Int2String7 implements Int2String {public synchronized String int2String(int intValue) {int position = 10;if (intValue < 0) {intValue = -intValue;buf[0] = '-';}while (true) {int leftValue = intValue % 10;buf[position] = (char) (leftValue + '0');intValue = (intValue >>1)/5;if (intValue == 0) {break;} else {position--;}}return new String(buf, position, buf.length - position);}}

打印：

1the time is 616, strValue = 100121312

总结 ：

整形转字符串，减少循环次数可以很明显提升运算速度。乘法运算比除法运算效率更高。尽量使用乘法代替除法。

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