Jian Zhi Offer 05. Replace spaces
Please implement a function to replace each space in the string s with "%20".
Method 1: Use StingBuffer
public static String replaceSpace(String s) {
//new StringBulider（速度快，线程不安全）或者StringBuffer填充
StringBuilder sb = new StringBuilder();
char[] arry = s.toCharArray();
for (int j = 0; j < arry.length; j++) {
if (arry[j] == ' ') {
sb.append("%20");
} else sb.append(arry[j]);
}
// 同String.valueOf(sb);
return sb.toString();
}
1. Convert String type data to char type array,
2. Use StingBuffer to create a new modifiable array
3. Traverse the array one by one, insert characters that are not spaces directly, and insert characters that are spaces with %20
4. Finally Convert data to String type
Method 2: Use the API directly
public static String replaceSpaceTwo(String s) {
String value = s.replaceAll(" ","%20");
return value;
}
Using replace should convert a single character, but changing replaceAll to replace can also perform the same result.
Method 3: Insert in reverse order
1. First traverse the string and record the number of spaces, and create a character with length + number of spaces*2 Array
2. Traverse the string from the back
3. If the character is a non-space, insert it as the original character, pay attention to the number of the insertion position -1
3. If the character is a space, insert %20 one by one, and each character -1
public static String replaceSpaceThree(String s){
int count = 0;
int oldlen = s.length() - 1;
// 旧数组长度-1，表示最后一个数的下标
char[] sArry = s.toCharArray();
for(int i =0; i < s.length(); i++){
if(sArry[i] == ' '){
count ++;
}
}
char[] value = new char[(s.length() + count *2)];
// 新的数组长度-1，表示最后一个数的下标
int newlen = value.length-1;
for(int j = oldlen; j >= 0;j--){
if(sArry[j] == ' '){
value[newlen--] = '0';
value[newlen--] = '2';
value[newlen--] = '%';
}
else{
//--是先赋值后再-1
value[newlen--] = sArry[j];
}
}
return String.valueOf(value);
}
Note: Insert in reverse order, so %20 of the insertion position must become 02%
The idea is not difficult. The blogger's first idea was to insert in a later order, and later thought that the length of the new array was determined, why not insert it in a positive order? It turns out that as long as two variables are used to control the subscript when inserting the array, it can be achieved. The blogger first used a subscript to control the new array subscripts +1 and +2, but it is difficult to determine the subscript assigned to j when there is no space in the subsequent sequence! Not much to say here!
Method 4: Positive sequence insertion
public static String replaceSpaceFour(String s){
int count = 0;
int oldlen = s.length() - 1;
// 旧数组长度-1，表示最后一个数的下标
char[] sArry = s.toCharArray();
for(int i =0; i < s.length(); i++){
if(sArry[i] == ' '){
count ++;
}
}
char[] value = new char[(s.length() + count *2)];
// 新的数组下标从0开始
int newlen = 0;
for(int j = 0; j <= oldlen;j++){
if(sArry[j] == ' '){
value[newlen++] = '%';
value[newlen++] = '2';
value[newlen++] = '0';
}
else{
//--是先赋值后再-1
value[newlen++] = sArry[j];
}
}
return String.valueOf(value);
}
The organization of these two methods wastes a lot of my time, because the official Jianzhi offer is the following method. First, a space of 3 times the length is opened up, and then inserted one by one, and finally the first size bit used is output , I personally feel that a lot of space was wasted during the first development, so I have been thinking about the above writing, it is an example that can be learned!
Method 5: Sword refers to offer
class Solution {
public String replaceSpace(String s) {
int length = s.length();
char[] array = new char[length * 3];
int size = 0;
for (int i = 0; i < length; i++) {
char c = s.charAt(i);
if (c == ' ') {
array[size++] = '%';
array[size++] = '2';
array[size++] = '0';
} else {
array[size++] = c;
}
}
String newStr = new String(array, 0, size);
return newStr;
}
}
In fact, observing the whole process, there is no difference, just the difference in the length of the array at the beginning!