http://www.iteye.com/topic/122472?page=1
需求:把字符串"Tdd is a software devolopment technology" 按照单词反转为
"technology devolopment software a is Tdd"
是看了这个帖子后一直有困惑。直接促成了我找了Kent Beck大的Test-Driven Development
By Example来看。自此进入TDD的世界。其实这题目和书后面的那个fibonacci的解法一样。
关键在于：“分解多项式” 且听我忽悠..

import junit.framework.TestCase;
public class NewStringReverserTest extends TestCase{
	//"a b" - "b a"
	public void testSimpleReverse(){
		assertEquals("b a",reverse("a b"));//从最简单的情况开始
	}

 	private String reverse(String string) {		
		return "b a";//最“直接”的实现，绿了吧- -！
	}
}

public class NewStringReverserTest extends TestCase{
	...
	private String reverse(String string) {		 	
		return "b"+" "+"a";//拆，为什么这么拆？ 往下看
	}
}

public class NewStringReverserTest extends TestCase{
	...
	private String reverse(String string) {	
                //要拆得有意义，注意描述= =
                //其实是一个一个拆的，每步都执行test，边绿边拆...小步快跑	 	
		String splitToken = " ";
		String preWord = "a";
		String postWord = "b";
                //下面其实是业务逻辑了，看后面的postWord被放在了前面——所谓的交换
		return postWord + splitToken + preWord;
	}
}

public class NewStringReverserTest extends TestCase{
	...
	private String reverse(String inputString) {	               
		String splitToken = " ";
                //实现切词逻辑
		String preWord =    inputString.substring(0,inputString.indexOf(splitToken));
		String postWord = inputString.substring(inputString.indexOf(splitToken)+1);               
		return postWord + splitedToken + preWord;
	}
}

ok 目前为止似乎无法继续了。那么继续添加test描述需求："a b c"-"c b a"

public class NewStringReverserTest extends TestCase{
	//"a b" - "b a"
	public void testSimpleReverse(){
		assertEquals("b a",reverse("a b"));
                assertEquals("c b a",reverse("a b c"));//红了吧？- -
	}
        ...
}

Expected：c b a Actual:b c a
好吧 分析下：bc 反了，bc看做一个整体，再“交换”一次就ok了,即: reverse("b c")+"a"

        private String reverse(String inputString) {	               
		...          
		return reverse(postWord) + splitToken + preWord;
	}

还是会红，debug发现“切到剩一个词”了就不能切了嘛= =

        private String reverse(String inputString) {	
                int tokenIndex = inputString.indexOf(splitToken);
		if (tokenIndex == -1)
			return inputString;
		String prePart = inputString.substring(0,tokenIndex);
		String postPart = inputString.substring(tokenIndex + 1);               
		//描述为prePart，postPart 更符合语义，也是一步重构吧        
		return reverse(postPart)+ splitToken + prePart
	}

这下绿了，想想其实prePart其实是reverse(prePart)
另外把那个需求的test加上 看看符合需求否？

       public class NewStringReverserTest extends TestCase {
	// "a b" - "b a"
	public void testSimpleReverse() {
		assertEquals("b a",reverse("a b"));
		assertEquals("c b a",reverse("a b c"));
                assertEquals("Tdd is a software devolopment technology",reverse("technology devolopment software a is Tdd"));

	}

	private String reverse(String inputString) {
		String splitToken = " ";
		int tokenIndex = inputString.indexOf(splitToken);
		if (tokenIndex == -1)
			return inputString;
		String prePart = inputString.substring(0,tokenIndex);
		String postPart = inputString.substring(tokenIndex + 1);
		return reverse(postPart) + splitToken + reverse(prePart);
	}

	public void testReverseWithSingleWords() {
		assertEquals("a",reverse("a"));
	}

	public void testReverseWithSplitTokenIn() {
		assertEquals(" ",reverse(" "));
		assertEquals("   ",reverse("   "));
		assertEquals("   a",reverse("a   "));
		assertEquals("b a",reverse("a b"));
		assertEquals("c b a ",reverse(" a b c"));
		assertEquals("c b a",reverse("a b c"));
	}

}

最后总结："a b c"到"c b a" 原来是a b c->[b c] a->[c b] a 原文链接：https://www.f2er.com/javaschema/288072.html