替换空格 4

先遍历每个字符统计空格数 String.charAt()

   

根据空格数计算新的长度 newLength=oriLength+2*NumOfBlank

   

新建一个newLength长度的字符数组tempChars用于存放替换空格之后的字符数组

   

利用System.arraycopy函数将原字符串转换为数组拷贝到新的字符数组tempChars

   

从新字符数组末尾开始遍历,用一个index1指向新字符数组末尾,用一个index2指向原字符数组末尾所在新字符数组的位置,将原字符数组逐个复制到新的字符数组的位置,遇到空格,替换,此时index1动,index2不动

   

《替换空格 4》

复杂度O(n),如果是从前往后复制的话复杂度O(n^2)

   

package replaceBlank;

   

public class ReplaceBlank {

public static int getBlankNum(String testString) {

int count = 0;

for (int i = 0; i < testString.length(); i++) {

String tempString = String.valueOf(testString.charAt(i));

if (tempString.equals(” “)) {

count++;

}

}

   

return count;

}

   

public static void main(String[] args) {

// TODO Auto-generated method stub

String string = “we are here”;

replaceBlank(string);

}

   

static void replaceBlank(String string){

if (string==null) {

return;

}

int orgLength=string.length();

int numOfBlank=0;

for (int i = 0; i < string.length(); i++) {

String tempString = String.valueOf(string.charAt(i));

if (tempString.equals(” “)) {

numOfBlank++;

}

}

int newLength=orgLength+2*numOfBlank;

char[] tempChars=new char[newLength];

System.arraycopy(string.toCharArray(), 0, tempChars, 0, string.length());

System.out.println(“orgLength:”+orgLength+”\n”+”numOfBlank:”+numOfBlank);

int indexOfOrgChars=orgLength-1;

int indexOfNewChars=newLength-1;

while (indexOfOrgChars>=0&&indexOfNewChars!=indexOfOrgChars) {

if (tempChars[indexOfOrgChars]==’ ‘) {

tempChars[indexOfNewChars–]=’%’;

tempChars[indexOfNewChars–]=’2′;

tempChars[indexOfNewChars–]=’0′;

}else {

tempChars[indexOfNewChars–]=tempChars[indexOfOrgChars];

}

indexOfOrgChars–;

}

System.out.println(tempChars);

}

}

    原文作者:keedor
    原文地址: https://www.cnblogs.com/keedor/p/4379086.html
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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