LCOF 05. 替换空格

1. 问题

请实现一个函数，把字符串 s 中的每个空格替换成"%20"。

示例 1：

输入：s = "We are happy."
输出："We%20are%20happy."

限制：

• 0 <= s 的长度 <= 10000

2. 解法

2.1 Java

class Solution {
    public String replaceSpace(String s) {
        char[] nums = new char[s.length() * 3];
        int index = 0;

        for (int i=0;i<s.length();i++) {
            if(s.charAt(i) != ' ') {
                nums[index++] = s.charAt(i);
            } else {
                nums[index++] = '%';
                nums[index++] = '2';
                nums[index++] = '0';
            }
        }

        String res = new String(nums, 0, index);
        return res;

    }
}

2.2 Kotlin

class Solution {
    fun replaceSpace(s: String): String {
        val nums = CharArray(s.length * 3)
        var index = 0

        for (i in 0..s.length-1) {
            if (s[i] != ' ') {
                nums[index++] = s[i]
            } else {
                nums[index++] = '%'
                nums[index++] = '2'
                nums[index++] = '0'
            }
        }

        return String(nums, 0, index)
    }
}

3. 参考

• https://leetcode-cn.com/problems/ti-huan-kong-ge-lcof/

• https://leetcode-cn.com/problems/ti-huan-kong-ge-lcof/solution/mian-shi-ti-05-ti-huan-kong-ge-by-leetcode-solutio/

4. 笔记