谁能帮我解释一下这个递归方法
[code=Java][/code]//后缀转中缀表达式end为postfix.length()-1,priority参数为1
public static String postToCenter(String[] postfix, int end, int priority) {
//判断最后一位是不是数字,是则返回
if (end == 0 || Character.isDigit(postfix[end].charAt(postfix[end].length() - 1)))
return postfix[end];
//不是数字就是运算符
StringBuffer buffer = new StringBuffer();
//给每种运算符付一个优先级（+ < - < * </）
char s = postfix[end].charAt(0);
if (s == '+')
s = 2;
else if (s == '-')
s = 3;
else if (s == '*')
s = 6;
else if (s == '/')
s = 7;
int k = end - 1;
for (int j = 0; k >= 0; k--) {
if (Character.isDigit(postfix[k].charAt(postfix[k].length() - 1)))
j++;
else
j--;
if (j == 1)
break;
}
if (s < priority)
buffer.append('(');
buffer.append(postToCenter(postfix, k - 1, s - s % 2));
buffer.append(postfix[end--]);
buffer.append(postToCenter(postfix, end, s + s % 2));
if (s < priority)
buffer.append(')');
System.out.println(new String(buffer));
return new String(buffer);
}

------解决方案--------------------

Java code

//后缀转中缀表达式end为postfix.length()-1,priority参数为1
public static String postToCenter(String[] postfix, int end, int priority) {
//判断最后一位是不是数字,是则返回
if (end == 0 || Character.isDigit(postfix[end].charAt(postfix[end].length() - 1)))
return postfix[end];
//不是数字就是运算符
StringBuffer buffer = new StringBuffer();
//给每种运算符付一个优先级（+ < - < * </）
char s = postfix[end].charAt(0);
if (s == '+')
s = 2;
else if (s == '-')
s = 3;
else if (s == '*')
s = 6;
else if (s == '/')
s = 7;
int k = end - 1;
for (int j = 0; k >= 0; k--) {
if (Character.isDigit(postfix[k].charAt(postfix[k].length() - 1)))
j++;
else
j--;
if (j == 1)
break;
}
if (s < priority)
buffer.append('(');
buffer.append(postToCenter(postfix, k - 1, s - s % 2));
buffer.append(postfix[end--]);
buffer.append(postToCenter(postfix, end, s + s % 2));
if (s < priority)
buffer.append(')');
System.out.println(new String(buffer));
return new String(buffer);
}

------解决方案--------------------

要理解递归  你得先理解递归