几天前，我遇到了一个类似的问题：
我们有两个字符串A和B，它们具有相同的超级字符集。我们需要更改这些字符串以获得两个相等的字符串。在每次移动中，我们都可以执行以下操作之一：
1-交换一个字符串的两个连续字符 2-交换字符串的第一个和最后一个字符
可以对任一字符串执行移动。 为获得两个相等的字符串，我们需要的最小移动次数是多少？ 输入格式和约束： 输入的第一行和第二行包含两个字符串A和B.保证它们的字符相等的超集。 1 <=长度（A）=长度（B）<= 2000 所有输入字符都在'a'和'z'之间
输出格式： 将最小移动次数打印到输出的唯一行
示例输入： AAB BAA
示例输出： 1
说明： 交换字符串aab的第一个和最后一个字符，将其转换为baa。这两个字符串现在相等。
我一直在尝试不同的算法，但没有成功。我总是做一些让他们平等的事情，但从来没有做过最小的动作。
编辑：我的伪代码是：
swap=>moves++
begin
if(A[0]!=B[0]){
if(A[0]==B[length-1]){
swap(B[0], B[length-1]);
}
else if(B[0]==A[length-1]){
swap(A[0],A[length-1]);
}
}
for(int i=0;i<A.length();i++){
if(A[i]==B[i]){
continue;
}
for(int j=i+1;j<A.length();j++){
if(A[i]==B[j]){
for(int k=j;k>i;k--){
swap(B[k],B[k-1]);
}
break;
else if(B[i]=A[j]){
for(int k=j;k>i;k--){
swap(A[k],A[k-1]);
}
break;
}
}if(A==B){
break;
}
对于一些不清楚的人，我的问题是如何让它发挥作用，任何想法，任何事情。因为此刻，我很无能为力。
编辑： 所以这是我迄今为止所做的最好的主意：
public static void swap(char aChar, char bChar) {
char cChar;
cChar = aChar;
aChar = bChar;
bChar = cChar;
}
public static void main(String[] args) {
int moves = 0;
Scanner scanner = new Scanner(System.in);
System.out.println("Enter the string A:");
while (!scanner.hasNext("[A-Za-z]+")) {
System.out
.println("Nope, that's not it! Enter the value from \"a\" to \"z\"."
+ "Enter your string A again!");
scanner.nextLine();
}
String a = scanner.nextLine();
a.toLowerCase();
System.out.println("Enter the string B:");
while (!scanner.hasNext("[A-Za-z]+")) {
System.out
.println("Nope, that's not it! Enter the value from \"a\" to \"z\"."
+ "Enter your string B again!");
scanner.nextLine();
}
String b = scanner.nextLine();
b.toLowerCase();
scanner.close();
char[] aChar = a.toCharArray();
char[] bChar = b.toCharArray();
if ((a.length() >= 1 && a.length() <= 2000) && a.length() == b.length()) {
if (aChar[0] != bChar[0]) {
if (aChar[0] == bChar[bChar.length - 1]) {
swap(bChar[0], bChar[bChar.length - 1]);
moves++;
} else if (bChar[0] == aChar[aChar.length - 1]) {
swap(aChar[0], aChar[bChar.length - 1]);
moves++;
}
}
for (int i = 0; i < aChar.length; i++) {
if (aChar[i] == bChar[i]) {
continue;
}
for (int j = i + 1; j < aChar.length; j++) {
if (aChar[i] == bChar[j]) {
for (int k = j; k > i; k--) {
swap(bChar[k], bChar[k - 1]);
moves++;
}
break;
} else if (bChar[i] == aChar[j]) {
for (int k = j; k > i; k--) {
swap(aChar[k], aChar[k - 1]);
moves++;
}
break;
}
}
if (aChar == bChar) {
break;
}
}
System.out.println("Minimum moves: " + moves);
}
}
}
问题在于它没有给出最小的移动次数。可悲的是。 ：）
将尝试添加更多内容添加到此代码中，它可能会起作用，但就目前而言，这就是我所拥有的......
答案 0 :(得分：0)
public int findMinimumStringMovement() {
int moves = 0;
char[] aChar = a.toCharArray();
char[] bChar = b.toCharArray();
char cChar;
if (aChar != bChar) {
for (int i = 0; i < aChar.length; i++) {
for (int j = i + 1; j < aChar.length; j++) {
if (aChar[i] != bChar[i]) {
/*
* It checks if some character from the array of aChar
* is same as other character from the array of B and if
* it's j less then the length of the array. If it's
* true, then swap the characters and count moves
*/
if (aChar[j] == bChar[i] && j < aChar.length) {
for (int k = j; k > i; k--) {
cChar = aChar[k];
aChar[k] = aChar[k - 1];
aChar[k - 1] = cChar;
moves++;
}
/*
* In other case, if the last character of array
* aChar same as the some character of bChar and
* vice versa, then we should check if the i is
* equal to 0, in that case we swap 1st and last
* character of array and count as 1 move else if i
* value is less then the value of length of array
* divided with 2 then it swaps that character to
* the first one and then swaps with last and counts
* the moves.
*/
} else if (aChar[aChar.length - 1] == bChar[i]) {
if (i == 0) {
cChar = aChar[aChar.length - 1];
aChar[aChar.length - 1] = aChar[i];
aChar[i] = cChar;
moves++;
} else if (i < (aChar.length - 1) / 2) {
for (int k = i; k > 0; k--) {
cChar = aChar[k];
aChar[k] = aChar[k - 1];
aChar[k - 1] = cChar;
moves++;
}
cChar = aChar[i];
aChar[i] = aChar[aChar.length - 1];
aChar[aChar.length - 1] = cChar;
}
} else if (bChar[bChar.length - 1] == aChar[i]) {
if (i == 0) {
cChar = bChar[bChar.length - 1];
bChar[bChar.length - 1] = bChar[i];
bChar[i] = cChar;
moves++;
} else if (i < (aChar.length - 1) / 2) {
for (int k = i; k > 0; k--) {
cChar = aChar[k];
aChar[k] = aChar[k - 1];
aChar[k - 1] = cChar;
moves++;
}
cChar = aChar[i];
aChar[i] = aChar[aChar.length - 1];
aChar[aChar.length - 1] = cChar;
moves++;
}
/*
* And the last case is if there is no other option,
* then we asks if some characters in array with
* positions i and j are different and if the j
* value is less then length of the array and do the
* swap.
*/
} else {
if (aChar[j] != aChar[i] && j < aChar.length) {
if (aChar[j] == bChar[j]) {
cChar = bChar[j];
bChar[j] = bChar[i];
bChar[i] = cChar;
moves++;
}
cChar = aChar[j];
aChar[j] = aChar[i];
aChar[i] = cChar;
moves++;
} else if (bChar[j] != bChar[i] && j < aChar.length) {
if (aChar[j] == bChar[j]
&& aChar[j] != aChar[i]) {
cChar = aChar[j];
aChar[j] = aChar[i];
aChar[i] = cChar;
moves++;
}
cChar = bChar[j];
bChar[j] = bChar[i];
bChar[i] = cChar;
moves++;
}
}
/*
* At the end, if arrays are equal, then it is done.
*/
if (aChar == bChar) {
break;
}
}
}
}
}
return moves;
}
我希望你能发现它有帮助的人。很抱歉花了这么长时间才发布。 最好的问候。