[Offer] [5] [space] Alternatively

• Title Description
• Ideas analysis
• Java code
• Code link

Title Description

　　 Please implement a function, each of the spaces in the replacement string as "20%."
For example, enter "We are happy.", The output "We% 20are% 20happy." .

Ideas analysis

1. Replace implemented in Java characters can use Stringbuilt-in functionsreplaceAll
2. Consider the function without implementation in Java: a function of cattle exercises off the net as defined in the parameters for the StringBuffer, the return value is String.
3. The whole idea is to calculate the length of a space after the replacement string, two indexes, old index point to the original string end , the new index after the variable-length string pointed end , the old index forward scan starts from the value of the string in order to assign a new index position, if you encounter a space, the new index in turn assigned to 20%, continue scanning. .

Java code

public class Offer005 {
    public static void main(String[] args) {
        StringBuffer str = new StringBuffer("We are no happy");
        System.out.println(Offer005.replaceSpace(str));

    }
    public static String replaceSpace(StringBuffer str) {
        return Solution2(str);
    }
    private static String Solution1(StringBuffer str) {
        if (str == null) {
            throw new IllegalArgumentException("参数非法!");
        }
        int length = str.length();
        int indexOfOrig = length - 1;// 旧的索引位置指向 旧的字符串末尾
        // 获取替换空格后的总长度
        for (int i = 0; i < str.length(); i++) {
            if (str.charAt(i) == ' ') {
                length += 2; // 因为要把空格替换成"%20",每次出现空格就要将长度增加2
            }
        }
        str.setLength(length);// 设置新的str长度
        int indexOfNew = length - 1;// 新的索引位置指向 新的字符串末尾
        while (indexOfNew > indexOfOrig) {
            if (str.charAt(indexOfOrig) != ' ') {
                str.setCharAt(indexOfNew--, str.charAt(indexOfOrig));
            } else {
                str.setCharAt(indexOfNew--, '0');
                str.setCharAt(indexOfNew--, '2');
                str.setCharAt(indexOfNew--, '%');
            }
            indexOfOrig--;
        }
        return str.toString();
    }
    private static String Solution2(StringBuffer str) {
        return str.toString().replaceAll(" ", "%20");
    }
}

Code link

Offer to prove safety codes -Java