"Sword refers to offer" 04: Replace spaces

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你自己的代码如果超过6个月不看，再看的时候也一样像是别人写的。—— 小浩

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Replace spaces

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Title description

Please implement a function to replace every space in a string with %20. For example, when the string is We Are Happy, the replaced string is We%20Are%20Happy.

Original title display

solution

Solution 1:
Create StringBuilder, traverse the original string, and directly append to StringBuilder when encountering non-spaces, and append %20 to StringBuilder when encountering spaces.

public class Solution {
    /**
     * 将字符串中的所有空格替换为%20
     * @param str 字符串
     * @return 替换后的字符串
     */
    public String replaceSpace(StringBuffer str) {
        if (str == null || str.length() == 0) {
            return str.toString();
        }
        StringBuilder sb = new StringBuilder();
        int len = str.length();
        for (int i = 0; i < len; ++i) {
            char ch = str.charAt(i);
            sb.append(ch == ' ' ? "%20" : ch);
        }

        return sb.toString();
    }
}

Solution 2 [Recommendation]
First traverse the original string, and if a space is encountered, append any two characters, such as two spaces, at the end of the original string.

Use the pointer p to point to the end of the original string, q point to the end of the current string, p, q traverse from back to front, when p encounters a space, the position of q should be appended '02%' in turn, if it is not a space, directly append the point pointed to by p character.

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思路扩展：在合并两个数组（包括字符串）时，如果从前往后复制每个数字（或字符）需要重复移动数字（或字符）多次，那么我们可以考虑从后往前复制，这样就能减少移动的次数，从而提高效率。

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public class Solution {
    /**
     * 将字符串中的所有空格替换为%20
     * @param str 字符串
     * @return 替换后的字符串
     */
    public String replaceSpace(StringBuffer str) {
        if (str == null || str.length() == 0) {
            return str.toString();
        }

        int len = str.length();
        for (int i = 0; i < len; ++i) {
            if (str.charAt(i) == ' ') {
                // append 两个空格
                str.append("  ");
            }
        }

        // p 指向原字符串末尾
        int p = len - 1;

        // q 指向现字符串末尾
        int q = str.length() - 1;

        while (p >= 0) {
            char ch = str.charAt(p--);
            if (ch == ' ') {
                str.setCharAt(q--, '0');
                str.setCharAt(q--, '2');
                str.setCharAt(q--, '%');
            } else {
                str.setCharAt(q--, ch);
            }
        }

        return str.toString();

    }
}

Test case

1. The input string contains spaces (the space is at the first/last/middle of the string; the string has multiple consecutive spaces);
2. There are no spaces in the input string;
3. Special input test (string is a null pointer; string is an empty string; string has only one space character; there are multiple consecutive spaces in the string).

   Test site for this question

   

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