在给定的字符串中，我想找出最大子字符串的起始索引，它们全部由相同的字符组成，并且还计算主字符串中出现子字符串的次数。
ex：“aaakkkkbbkkkkk” 在这种情况下，子串“kkkkk”的计数为5，起始位置为9。
到目前为止我的代码：
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);
}
我的问题：无法存储计数器值，因为此计数器值是子字符串的实际出现次数，稍后我会将子字符串与计数器出现进行比较。
我也不符合我的逻辑。
答案 0 :(得分：2)
1.使用单独的变量来存储第一次出现的子串
2.制作将存储最后序列长度的其他计数器
3.新序列启动时重置计数器
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);
}
答案 1 :(得分：2)
还有一个解决方案：
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;
}
}
答案 2 :(得分：2)
我认为这对你有帮助， 我知道它很小，很容易看到每一个，...
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);
}
快乐编码.. @ All .....
答案 3 :(得分：0)
此解决方案仅找到一个可能的子字符串，可能有几个具有相同最大重复字符数的子字符串。所以这是我的解决方案，它将返回一个列表。
代码如下：
import java.util.List; 导入java.util.ArrayList;
公共类SubString {
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);
}
}
}
//此示例字符串的输出为： 最长长度：2是字符：1开始于：3 最长的长度：2是字符：2开始于：6 最长的长度：2是字符：3始于：9