URL化

{URL化}https://leetcode-cn.com/problems/string-to-url-lcci/submissions/

1. 我第一次做法

   public String replaceSpaces(String s, int length) {
           StringBuilder result = new StringBuilder();
           char[] charArr = s.toCharArray();
           for (int i = 0; i < length; i++) {
               if (charArr[i] == ' ') {
                   result.append("%20");
               } else {
                   result.append(charArr[i]);
               }
           }
           return result.toString();
       }
   

   Fail - 超时了

2. 参考评论区做法

   发现自己没理解对，题目的本意是用字符数组来解答

    public String replaceSpaces1(String s, int length) {
           char[] chars = s.toCharArray();
           int i = length - 1;
           int j = chars.length - 1;
   
           while (i >= 0) {
               if (chars[i] == ' ') {
                   chars[j--] = '0';
                   chars[j--] = '2';
                   chars[j--] = '%';
               } else {
                   chars[j--] = chars[i];
               }
               --i;
           }
   
           return String.valueOf(chars, j + 1, chars.length - 1 - j);
       }
   

3. 类似于解法2

   public String replaceSpace2(String s, int length) {
           char[] chars = s.toCharArray();
           char[] result = new char[s.length()];
           int newIndex = 0;
           for (int i = 0; i < length; i++) {
               if (chars[i] == ' ') {
                   result[newIndex++] = '%';
                   result[newIndex++] = '2';
                   result[newIndex++] = '0';
               }else {
                   result[newIndex++] = chars[i];
               }
           }
           return String.valueOf(result, 0, newIndex);
       }
   

   但是内存消耗比较多

4. 优化2的内存使用

       /**
        * 减少了一次字符数组的copy
        */
    public String replaceSpace3(String s, int length) {
           char[] result = new char[s.length()];
           int newIndex = 0;
           for (int i = 0; i < length; i++) {
               if (s.charAt(i) == ' ') {
                   result[newIndex++] = '%';
                   result[newIndex++] = '2';
                   result[newIndex++] = '0';
               }else {
                   result[newIndex++] = s.charAt(i);
               }
           }
           return String.valueOf(result, 0, newIndex);
       }
   

5. 在4的基础上，寻找优化空间

      /**
        * while 替换 for 发现内存使用少了点
        */
       public String replaceSpace4(String s, int length) {
           char[] result = new char[s.length()];
           int newIndex = 0;
           int i = 0;
           while (i < length){
               if (s.charAt(i) == ' ') {
                   result[newIndex++] = '%';
                   result[newIndex++] = '2';
                   result[newIndex++] = '0';
               }else {
                   result[newIndex++] = s.charAt(i);
               }
   
               ++i;
           }
           return String.valueOf(result, 0, newIndex);
       }