Java 计数和所有字符都相同的最大字符串的起始索引
在给定的字符串中,我想找出由相同字符组成的最大子字符串的起始索引,并计算该子字符串在主字符串中出现的次数 例如:“aaakkkbkkk” 在这种情况下,子字符串“kkk”的计数为5,起始位置为9 到目前为止,我的代码是:Java 计数和所有字符都相同的最大字符串的起始索引,java,jakarta-ee,Java,Jakarta Ee,在给定的字符串中,我想找出由相同字符组成的最大子字符串的起始索引,并计算该子字符串在主字符串中出现的次数 例如:“aaakkkbkkk” 在这种情况下,子字符串“kkk”的计数为5,起始位置为9 到目前为止,我的代码是: String str = "aaakkkkbbkkkkk"; int count = 0; //converting string into character array char[] vals = str.toCharArray(); for(int i=0; i &l
String str = "aaakkkkbbkkkkk";
int count = 0;
//converting string into character array
char[] vals = str.toCharArray();
for(int i=0; i < vals.length; ){
for(int j=i+1; j<vals.length; j++){
//if value match then increment counter
if(vals[i]== str.charAt(j)){
counter++;
}
//else break from inner loop
break; //break from inner loop
}
//assign the index value of j to the i to start with new substring
i = vals.indexOf(j);
}
String str=“aaakkkbkk”;
整数计数=0;
//将字符串转换为字符数组
char[]vals=str.toCharArray();
对于(int i=0;i 对于(int j=i+1;j1),使用单独的变量存储子字符串的第一次出现
2.制作将存储在最后一个序列长度的其他计数器
3.在新序列开始时重置计数器
String str = "aaakkkkbbkkkkk";
int count = 0;
int firstOccurence = 0;
int prevSequenceLength = 0;
//converting string into character array
char[] vals = str.toCharArray();
for(int i=0; i < vals.length; ){
for(int j=i+1; j<vals.length; j++){
//if value match then increment counter
if(vals[i]== str.charAt(j)){
counter++;
}
else
{
//set the prevSequenceLength to counter value if it is less than counter value
if(prevSequenceLength<counter)
{
prevSequenceLength=counter;
//make firstOccurence to j - counter
firstOccurence = j-counter;
}
//first reset the counter to 0;
counter = 0;
//else break from inner loop
break; //break from inner loop
}
//assign the index value of j to the i to start with new substring
i = vals.indexOf(j);
}
String str=“aaakkkbkk”;
整数计数=0;
int firstoccurrence=0;
int-prevSequenceLength=0;
//将字符串转换为字符数组
char[]vals=str.toCharArray();
对于(int i=0;i
class Sample {
static int max = 0, maxStartPos = -1;
public static void main(String[] args) {
search("aaakkkkbbkkkkk".toCharArray());
System.out.println(String.format("max len: %d, start pos: %d", max, maxStartPos));
search("xxxxxh".toCharArray());
System.out.println(String.format("max len: %d, start pos: %d", max, maxStartPos));
search("hxxxxxhyyyyyyh".toCharArray());
System.out.println(String.format("max len: %d, start pos: %d", max, maxStartPos));
}
private static void search(char[] chars) {
max = 0; maxStartPos = 0;
int len = 0, startPos = 0;
char start = chars[0];
for (int i = 0; i < chars.length; i++) {
if (chars[i] == start) {
len++;
} else {
if (len > max) {
updateMaxResults(len, startPos);
}
startPos = i;
len = 1;
start = chars[i];
}
}
if (len > max) {
updateMaxResults(len, startPos);
}
}
private static void updateMaxResults(int len, int startPos) {
max = len;
maxStartPos = startPos;
}
}
类示例{
静态int max=0,maxStartPos=-1;
公共静态void main(字符串[]args){
搜索(“aaakkkbkkk.toCharArray());
System.out.println(String.format(“max len:%d,start pos:%d”,max,maxStartPos));
搜索(“xxxxx h”.toCharArray());
System.out.println(String.format(“max len:%d,start pos:%d”,max,maxStartPos));
搜索(“hxxxxxhyyyyyyyh.toCharArray());
System.out.println(String.format(“max len:%d,start pos:%d”,max,maxStartPos));
}
专用静态无效搜索(字符[]字符){
max=0;maxStartPos=0;
int len=0,startPos=0;
char start=chars[0];
for(int i=0;i最大值){
更新MaxResults(len、startPos);
}
startPos=i;
len=1;
开始=字符[i];
}
}
如果(长度>最大值){
更新MaxResults(len、startPos);
}
}
私有静态void updateMaxResults(int len、int startPos){
max=len;
maxStartPos=startPos;
}
}
我想这会对你有所帮助,,
我知道它很小,很容易看到每一个人
public static void main(String[] args) {
String str = "aaakkkkkkkbbkkkkk";
char[] var = str.toCharArray();
for(int i=0; i<var.length;i++)
System.out.print(" "+var[i]);
int final_index=0; // this is for final index what we want
int max_size=0; // for maximum no. of time the same char repeats continiously..
int size=0; // this for finding size of sub string..
char pre_char=var[0]; // pre_char is used check with present char in the array where the i position is shifted..
// here is the loop..
// we don't need much loops to this
// while we are reading , the comparison is also done parallely..
for(int i=1;i<var.length;i++) // here i starts from 1 because 0th char is shfted into pre_char
//and now comparion is starts from next char ie from 1th position
{
// System.out.println("i=="+i+" size "+size+" prechar== "+pre_char+ " var[i] ="+var[i]+" max size=="+max_size);
// System.out.println("..........................");
if(pre_char==var[i]) // here condition is checking. if its same with previous char, its means same char is occur again..
{
size+=1;
}else{ // else means the sub string is has different char
if(max_size<size) // now check for whether any maximum size is occured priviously or not..
{
max_size=size;
final_index=i-size;
}
size=0;
}
pre_char=var[i];
}
// now this for final
// its means if the max sub string is at the ending position, the above for loop breaks at the last element
// in this case we check if the last sub string is max or not..
if(max_size<size)
{
max_size=size;
final_index=var.length-size;
}
// then this is the final what we wana,,,,,
System.out.print("Max size is "+(max_size+1)+" index is "+final_index);
}
publicstaticvoidmain(字符串[]args){
String str=“aaakkkkbkkk”;
char[]var=str.toCharArray();
对于(int i=0;i此解决方案只找到一个可能的子字符串,可能有几个子字符串具有相同的最大重复字符数。因此,我的解决方案将返回一个列表
代码如下:
导入java.util.List;
导入java.util.ArrayList
公共类子串{
public static void main(String[] args) {
String a1 = "12311m22n33";
char[] arrays = (a1+"\n").toCharArray();
//value is char +"_" + start
List<String[]> lists = new ArrayList<>();
int len = 0;
int start = -1;
char prev = '\n';
int position= -1;
int maxLen = 0;
for (char a: arrays) {
position++;
if (a != prev) {
if (len > maxLen ) {
lists.clear();
String[] sq = {prev+"", start+""};
lists.add(sq);
maxLen = len;
} else if ( len == maxLen && len != 0) {
String[] sq = {prev+"", start+""};
lists.add(sq);
}
prev = a;
len = 1;
start = position;
} else {
len++;
}
}
for(String[] a: lists) {
String s = a[0];
String st = a[1];
System.out.println("The longest length: " + maxLen + " is char: " + s + " start at : " + st);
}
}
publicstaticvoidmain(字符串[]args){
字符串a1=“12311m22n33”;
字符[]数组=(a1+“\n”).toCharArray();
//值为char+“”+start
List lists=新建ArrayList();
int len=0;
int start=-1;
char prev='\n';
int位置=-1;
int maxLen=0;
for(字符a:数组){
位置++;
如果(a!=上一个){
如果(len>maxLen){
lists.clear();
字符串[]sq={prev+“”,start+“”};
列表。添加(sq);
maxLen=len;
}else如果(len==maxLen&&len!=0){
字符串[]sq={prev+“”,start+“”};
列表。添加(sq);
}
prev=a;
len=1;
开始=位置;
}否则{
len++;
}
}
对于(字符串[]a:列表){
字符串s=a[0];
字符串st=a[1];
System.out.println(“最长长度:“+maxLen+”是字符:“+s+”起始于:“+st”);
}
}
}
//此示例字符串的输出为:
最长长度:2是字符:1,从:3开始
最长的长度:2是字符:2,从:6开始
最长的长度:2是字符:3从:9开始让我给点时间来分析你的代码。一旦我满意了,我会给你打分。谢谢尼古拉。结果是:max len:5,start pos:9 max len:5,start pos:0 max len:6,start pos:7我只需要:max len:5,start pos:9。这只是一系列测试来证明代码可以处理输入中最长的子串:在输入开始时,在中间开始,在结尾处。谢谢。如果您在上面加上注释,我将非常高兴。我正在学习编程基础。您的注释代码将被添加。好的,我用注释改变我的答案来理解逻辑……@ Devendra Singhi认为评论是超过实际代码..@Devendra Singh