我有类LString,它模仿标准的Java String和StringBuilder类。我无法开始实现替换方法来替换子字符串的字符" LString"用" lStr"中的字符。我的替换方法是我的代码的底部。
如果if = = end,则插入" lStr"在LString的给定位置。如果start == end == this.length(),它会在此LString的末尾附加lStr。
public class LString {
node front;
int length;
LString next;
// create node class
public class node {
char data;
node next;
public node(char newData) {
data = newData;
}
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();
}
//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
return null;
} // end class
/*
// create new LString
LString newLString = new LString();
node node = this.front;
// to copy
node copy = new node(node.data);
// for node to replace
node replace = lStr.front;
if (start == end) {
LString newLStr = new LString();
return newLStr;
}
if (start == end && start == this.length()) {
LString newLStr = new LString();
return newLStr;
*/
}
/*
//append / prepend methods for linked list
//append
public void append(int data){
if (front == null){
front = new node(data);
end = front;
}
else {
end.next = new node(data);
end = end.next;
}
length++;
}
//prepend
public void prepend(int data){
if (front == null){
end = new node(data);
front = end;
}
else {
front = new node(data,front);
}
length++;
}
*/
replace之前的方法返回一个新的LString,它是该LString的子字符串。感谢您的帮助,谢谢。
答案 0 :(得分:1)
如果你有一个append和subsequence / substring方法,你可以这样做:
public LString replace(int start, int end, LString replacement){
//Do exception checks for start/end values here
LString replacedLs = new LString();
replacedLs.append(substring(0, start));
replacedLs.append(replacement);
replacedLs.append(substring(end, this.length));
return replacedLs;
}
答案 1 :(得分:1)
asiew的答案听起来很棒(更简单更好,更安全),但是我们可以通过编写insert和remove方法使事情复杂化,然后替换可能如下所示:
1)插入 2)删除 3)胜利
每种方法的一个(可能)次优解决方案可能如下所示:
private int insert(LString s, int index ) {
node
n1 = front,
n2 = null;
for (int i = 0; i < index-1; i++) {
n1 = n1.next;
}
n2 = n1.next;
if (n2 == null) return -1;
int inserted = 0;
for (char c : s.toString().toCharArray()) {
n1.next = new node(c);
n1 = n1.next;
inserted++;
}
n1.next = n2;
return inserted;
}
private int remove(int from, int to) {
node n1 = front,n2 = null;
for (int i = 0; i < to+1; i++) {
if (i<from) {
n1=n1.next;
} else if (i==from) n2=n1.next;
else n2=n2.next;
}
n1.next = n2;
return -1; // or whatever
}
public LString replace(int start, int end, LString lStr) {
int inserted = insert(lStr,start);
int removed = remove(inserted+start-1,inserted+end-1);
return this;
}
当然我们应该在这里和那里检查nulls无效参数。