Java 如何递归执行蛮力方法
我正在尝试使用蛮力创建一个递归算法。假设此算法比较整数,如果比较的整数不相似,则“编辑距离”。示例:3254应该是2345。这里的编辑距离是2[(3,2)(5,4)] 我的代码可以编译,但没有任何输出。如果有人能帮我排除故障,我将不胜感激Java 如何递归执行蛮力方法,java,algorithm,permutation,Java,Algorithm,Permutation,我正在尝试使用蛮力创建一个递归算法。假设此算法比较整数,如果比较的整数不相似,则“编辑距离”。示例:3254应该是2345。这里的编辑距离是2[(3,2)(5,4)] 我的代码可以编译,但没有任何输出。如果有人能帮我排除故障,我将不胜感激 public static int measureEditDistance(int[] rankings) { int editDistance = 0; int R = rankings.length; // You need to
public static int measureEditDistance(int[] rankings) {
int editDistance = 0;
int R = rankings.length;
// You need to write this method.
// Add your logic here to compute the editDistance
for (int m = 0; m < R; m++) {
for (int i = m + 1; i < R; i++) {
for (int j = i + 1; j < R; j++) {
if (rankings[m] + rankings[i] + rankings[j] == 0) {
editDistance++;
}
}
}
}
return editDistance;
}
公共静态int-measureditdistance(int[]排名){
int editDistance=0;
int R=排名。长度;
//您需要编写此方法。
//在此处添加逻辑以计算editDistance
对于(int m=0;m
真正的暴力方法需要找到给定数字的所有排列。下面是实现这一点的方法(递归):
所以我们可以检查一下。我们必须将permute(int[])
的代码更改为:
int permute(int[] arr, int[] rankings){
int distancevar = 0;
distance var = permute(arr,rankings,0,arr.length-1);
return distancevar;
}
然后我们必须将permute(int[],int,int)
的返回类型更改为int
,以便返回我们发现为真的距离。这将使代码如下所示:
int permute(int[] arr, int[] rankings, int i, int n) {
int j;
int ans = 0;
bool isAns = false;
if (i == n)
System.out.println(Arrays.toString(arr));
else {
for (j = i; j <= n; j++) {
swap(arr, i, j);
isAns = check(arr, rankings);
if(isAns){
**here you would call your distance check on the two arrays,
which I am still unsure of what you actually are doing here**
ans = *some distance calc*;
return ans;
}
ans = permute(arr,rankings, i + 1, n);
if (ans == 0){
swap(arr, i, j); // backtrack
isAns = check(arr, rankings);
if(isAns){
**here you would call your distance check on the two arrays,
which I am still unsure of what you actually are doing here**
ans = *some distance calc*;
return ans;
}
}
}
return ans;
}
解释
注意permute函数如何调用自身并返回
ans
其中ans=distance\u value
。然后通过所有函数调用传回该值。这是递归。请将代码粘贴为文本,而不是文本image@Mureinik但你为什么要将图像内联?那是错误的教训。截图一文不值。请将代码复制到您的帖子中,删除图像。你是否觉得拍摄截图、上传并复制链接比直接复制代码更快?你看不到任何输出,因为它没有打印出来。谷歌将了解如何在Java@andrew在我可以帮助回答之前,是否需要使用TRUE bruteforce(如果答案是错误的,请放弃所有内容并尝试新的)?或者你能优化一下,一旦数字在正确的位置就不交换吗?@Andrew我希望这对朋友有帮助。如果代码不适合您,请告诉我,我会编辑或删除它。只需编辑它以修复逻辑中的缺陷。返回了错误的答案,以前每次都可能是0。
int permute(int[] arr, int[] rankings){
int distancevar = 0;
distance var = permute(arr,rankings,0,arr.length-1);
return distancevar;
}
int permute(int[] arr, int[] rankings, int i, int n) {
int j;
int ans = 0;
bool isAns = false;
if (i == n)
System.out.println(Arrays.toString(arr));
else {
for (j = i; j <= n; j++) {
swap(arr, i, j);
isAns = check(arr, rankings);
if(isAns){
**here you would call your distance check on the two arrays,
which I am still unsure of what you actually are doing here**
ans = *some distance calc*;
return ans;
}
ans = permute(arr,rankings, i + 1, n);
if (ans == 0){
swap(arr, i, j); // backtrack
isAns = check(arr, rankings);
if(isAns){
**here you would call your distance check on the two arrays,
which I am still unsure of what you actually are doing here**
ans = *some distance calc*;
return ans;
}
}
}
return ans;
}
bool check(int[] arr, int[] ans)
int same = 0
for(int i = 0; i < temp.length; i++){
if(temp[i] == ans[i])
same++;
if (same == 4)
return true;
return false;
int ans = permute(*your array to test*, *final array desired*);
System.out.println(ans);