我有方法replace（）和append（）允许替换方法替换子字符串的字符＆＃34; LString＆＃34;与＃34; lStr＆＃34;中的人物在Linked List类LString中。但是我无法在我的替换类中解析我的append方法。
可能是因为我没有为节点构造函数提供LString参数。我不确定如何纠正这个问题。相关方法是我班级的最底层。
public class LString {
node front;
node end;
int length;
LString next;
char letter;
// create node class
public class node {
char data;
node next;
public node(LString newData) {
LString data;
}
public node(char newData, node newNext) {
data = newData;
next = newNext;
}
}
// LString constructor
// constructs LString object representing empty list of chars
public LString() {
this.length = 0;
this.front = null;
}
// Construct LString copy of original parameter
public LString(String original) {
//assign first char to front
node curr = this.front;
this.length = original.length();
// loop through
for (int i = 0; i < original.length(); i++) {
if (i == 0) {
front = new node(original.charAt(i), null);
curr = front;
} else {
curr.next = new node(original.charAt(i), null);
curr = curr.next;
}
}
}
// return length in this LString
// can use in LString constructor
public int length() {
return this.length;
}
//create and return string with contents of LString
public String toString() {
// use string builder to meet time limit
StringBuilder builder = new StringBuilder();
if (front == null) {
return "";
} else {
node curr = front;
while (curr != null) {
builder.append(curr.data);
curr = curr.next;
}
return builder.toString();
}
}
//compares string lists 0 if equal -1 if less, and 1 if greater
public int compareTo(LString anotherLString) {
//save lowest length of strings
int minLength;
// get front spots
node curr = this.front;
node otherCurr = anotherLString.front;
// get lengths
int thisString = length();
int otherString = anotherLString.length();
// get shortest length of 2 strings
if (thisString < otherString) {
minLength = thisString;
} else {
minLength = otherString;
}
//go through characters in each string and compare for lexicographic order
int iterate = 0;
while (iterate < minLength) {
char string1Char = curr.data;
char string2Char = otherCurr.data;
if (string1Char != string2Char) {
return string1Char - string2Char;
}
iterate++;
curr = curr.next;
otherCurr = otherCurr.next;
}
return thisString - otherString;
}
//Return true if LString represents the same list of characters as other
@Override
public boolean equals(Object other) {
if (other == null || !(other instanceof LString))
return false;
else {
// use compareTo to determine if strings are the same
LString otherLString = (LString) other;
if (compareTo(otherLString) == 0) {
return true;
} else {
return false;
}
}
}
//Return the char at the given index in this LString.
public char charAt(int index) {
int length = this.length();
// check for index out of bounds
if (index < 0 || index >= length) {
throw new IndexOutOfBoundsException();
}
// returns char at index
node curr = front;
for (int i = 0; i < index; i++) {
curr = curr.next;
}
return curr.data;
}
//Set the char at the given index in this LString to ch.
public void setCharAt(int index, char ch) {
// check for index out of bounds
int length = this.length();
if (index < 0 || index >= length) {
throw new IndexOutOfBoundsException();
}
// replaces char at index
node curr = front;
for (int i = 0; i < index; i++) {
curr = curr.next;
}
curr.next = new node(curr.data, curr.next);
curr.data = ch;
}
//Returns a new LString that is a sub-string of this LString.
public LString substring(int start, int end) {
LString newLString = new LString();
// handle exceptions
if (start < 0 || start > end) {
throw new IndexOutOfBoundsException();
} else if (end > this.length()) {
throw new IndexOutOfBoundsException();
//return null in special case (empty LString)
} else if (start == end) {
//&& end == this.length()
return newLString;
} else {
node node = this.front;
for (int i = 0; i < start; i++) {
node = node.next;
// insert substring
}
node copy = new node(node.data);
newLString.front = copy;
for (int i = start + 1; i < end; i++) {
node = node.next;
copy = copy.next = new node(node.data);
}
return newLString;
}
}
// Replaces this characters in a sub-string of this LString with the characters in lStr.
public LString replace(int start, int end, LString lStr) {
// handle exceptions
if (start < 0 || start > end) {
throw new IndexOutOfBoundsException();
} else if (end > this.length()) {
throw new IndexOutOfBoundsException();
}
LString replacedLs = new LString();
replacedLs.append(substring(0, start));
replacedLs.append(lStr);
replacedLs.append(substring(end, this.length));
return replacedLs;
}
//append helper method
public void append(LString data) {
this.length++;
if (front == null) {
front = new node(data);
return;
}
node curr = front;
while (curr.next != null) {
curr = curr.next;
}
curr.next = new node(data);
}
}
感谢您的帮助，谢谢。
答案 0 :(得分：0)
看看你的node构造函数，它什么都不做：
public node(LString newData) {
LString data;
}
要使其工作，您应该将所有LString.front字段分配给当前对象，所以：
public node(LString newData) {
this.data = newData.front.data;
this.next = newData.front.next;
}