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参考技术AJAVA 中int类型转String类型的通常方法,有三种：
1、String.valueOf(int i)
2、Integer.toString(int i)
3、i+""; //i 为 int类型
是string类型的字符串跟上个int类型的i代表的字符串（先把i转换为string），int+string型就是先将int型的i转为string然后跟上后面的string。
三种方法效率排序为：
Integer.toString(int i) > String.valueOf(int i) > i+"";
java int转String所有方式的效率对比与深入解析
在java中，大家肯定都会遇到int类型转String类型的情形，知其然知其所以然，总结加分析一下，int类型转String类型有以下几种方式：
- a+”“
- String.valueOf(a)
- Integer.toString(a)
int a = 123456789;long start = System.currentTimeMillis();for (int i=0; i<100000; i++){ String m = a+"";}long end = System.currentTimeMillis();Log.e("time", "a+"" = " + (end - start));start = System.currentTimeMillis();for (int i=0; i<100000; i++){ String n = String.valueOf(a);}end = System.currentTimeMillis();Log.e("time", "String.valueOf(a) = " +(end-start));start = System.currentTimeMillis();for (int i=0; i<100000; i++){ String n = Integer.toString(a);}end = System.currentTimeMillis();Log.e("time", "Integer.toString(a) = " +(end-start));最后打印出来的执行时间：
E/time: a+"" = 257E/time: String.valueOf(a) = 140E/time: Integer.toString(a) = 159可以看到在效率上除了a+”“这种方式之外，其他两种方式的效率差不多，为什么呢？看源码！
先看看后两种方式的源码： String.valueOf(a)->Integer.toString(a)->IntegralToString.intToString(a)->convertInt(null, a)
Integer.toString(a)->IntegralToString.intToString(a)->convertInt(null, a)
可以看到String.valueOf是通过调用Integer.toString实现的，也难怪他们的效率如此接近。他们最后都会调用到convertInt函数中：
private static String convertInt(AbstractStringBuilder sb, int i) { boolean negative = false; String quickResult = null; if (i < 0) { negative = true; i = -i; if (i < 100) { if (i < 0) { // If -n is still negative, n is Integer.MIN_VALUE quickResult = "-2147483648"; } else { quickResult = SMALL_NEGATIVE_VALUES[i]; if (quickResult == null) { SMALL_NEGATIVE_VALUES[i] = quickResult = i < 10 ? stringOf("-", ONES[i]) : stringOf("-", TENS[i], ONES[i]); } } } } else { if (i < 100) { quickResult = SMALL_NONNEGATIVE_VALUES[i]; if (quickResult == null) { SMALL_NONNEGATIVE_VALUES[i] = quickResult = i < 10 ? stringOf(ONES[i]) : stringOf(TENS[i], ONES[i]); } } } if (quickResult != null) { if (sb != null) { sb.append0(quickResult); return null; } return quickResult; } int bufLen = 11; // Max number of chars in result char[] buf = (sb != null) ? BUFFER.get() : new char[bufLen]; int cursor = bufLen; // Calculate digits two-at-a-time till remaining digits fit in 16 bits while (i >= (1 << 16)) { // Compute q = n/100 and r = n % 100 as per "Hacker"s Delight" 10-8 int q = (int) ((0x51EB851FL * i) >>> 37); int r = i - 100*q; buf[--cursor] = ONES[r]; buf[--cursor] = TENS[r]; i = q; } // Calculate remaining digits one-at-a-time for performance while (i != 0) { // Compute q = n/10 and r = n % 10 as per "Hacker"s Delight" 10-8 int q = (0xCCCD * i) >>> 19; int r = i - 10*q; buf[--cursor] = DIGITS[r]; i = q; } if (negative) { buf[--cursor] = "-"; } if (sb != null) { sb.append0(buf, cursor, bufLen - cursor); return null; } else { return new String(cursor, bufLen - cursor, buf); }}分析一下，这个函数的工作主要可以分为这几步：
- 如果a为负数，将a变成正数，如果a还小于0，直接置为Integer.MIN_VALUE；如果a小于100，则直接使用TENS和ONES数组进行快速计算得出结果，加上’-"号，直接返回该结果。
- 如果a为正数并且小于100，直接使用TENS和ONES数组进行快速计算得出结果返回。
- 如果上面两步没有处理完，说明a是大于100的数字，无法直接使用TENS和ONES数组进行快速计算，处理方式就是2位为一步循环处理，每次将这两位使用TENS和ONES数组进行快速计算得出这两位的结果存在数组的相应位置，直到只剩一位；最后剩下的一位使用DIGITS数组得出16进制的结果放在最后，返回结果。
// Compute q = n/100 and r = n % 100 as per "Hacker"s Delight" 10-8 // Compute q = n/10 and r = n % 10 as per "Hacker"s Delight" 10-8 去看《Hacker’s Delight》的10-8章。
接着还有一个问题是TENS和ONES数组，直接看代码，一目了然：
/** TENS[i] contains the tens digit of the number i, 0 <= i <= 99. */private static final char[] TENS = { "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "1", "1", "1", "1", "1", "1", "1", "1", "1", "1", "2", "2", "2", "2", "2", "2", "2", "2", "2", "2", "3", "3", "3", "3", "3", "3", "3", "3", "3", "3", "4", "4", "4", "4", "4", "4", "4", "4", "4", "4", "5", "5", "5", "5", "5", "5", "5", "5", "5", "5", "6", "6", "6", "6", "6", "6", "6", "6", "6", "6", "7", "7", "7", "7", "7", "7", "7", "7", "7", "7", "8", "8", "8", "8", "8", "8", "8", "8", "8", "8", "9", "9", "9", "9", "9", "9", "9", "9", "9", "9"};/** Ones [i] contains the tens digit of the number i, 0 <= i <= 99. */private static final char[] ONES = { "0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "0", "1", "2", "3", "4", "5", "6", "7", "8", "9",};每个数组都是100的长度，都是用来处理0~99这100个数字，个位和十位的处理方式也很清楚。
从代码角度来看，这个算法在数字小于100的和大于100的处理方式是不一样的，小于100的快速计算法执行时间会远远短于大于100的方式，验证一下，将a变量修改为10：
E/time: i+"" = 199E/time: String.valueOf() = 7E/time: Integer.toString() = 6确实短了很多！！！
再来看看a+”“的方式，我承认这种方式我用的最多了，因为太简单了，java源码对’+’运算符进行了重载，源码我找不到啊，不过从网上找一些资料：
The Java language provides special support for the string concatenation operator ( + ), and for conversion of other objects to strings. String concatenation is implemented through the StringBuilder(or StringBuffer) class and its append method. String conversions are implemented through the method toString, defined by Object and inherited by all classes in Java. For additional information on string concatenation and conversion, see Gosling, Joy, and Steele, The Java Language Specification.
地址：http://docs.oracle.com/javase/6/docs/api/java/lang/String.html
可以看到，’+’运算符的主要方式是使用StringBuilder或者StringBuffer来实现的，类似于：
StringBuilder sb = new StringBuilder();sb.append("");sb.append(i);String strI = sb.toString();再来看看append的源码： StringBuffer.append->IntegralToString.appendInt(this, a)->convertInt(sb, i)
可以看到’+’运算符最后也是调用到了同一个函数，只不过第一个参数的sb不为null而已，所以已经很清楚了，’+’运算符的执行效率不高的原因应该就在之前的new StringBuilder等操作和之后的StringBuilder.toString等操作，反编译class文件也可以得出一样的结论：
http://stackoverflow.com/a/4105406。
所以a+”“的方式以后就少用一点了，效率不高，也显得不太专业。
经过几个星期的面试，发现自己的 java 基础确实很弱，所以以后要多多加强 java 基础，从这开始，加油~
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