LeetCode 每日一题 [40] 替换空格

LeetCode 替换空格 [简单]

请实现一个函数，把字符串 s 中的每个空格替换成"%20"。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/ti-huan-kong-ge-lcof

示例 1：

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

限制：

0 <= s 的长度 <= 10000

题目分析

解法1

使用Java的String的API String().replace("","");

解法2

将字符串转化为字符数组，然后遍历。符合条件则直接加入即可

代码实现

public class ReplaceSpace {
    public static void main(String[] args) {
        String s = "We are happy.";
        String s2 = "We are happy.";
        System.out.println(replaceSpace(s));
        System.out.println(replaceSpace2(s2));
    }

    public static String replaceSpace2(String s) {
        if (s == null || s.length() == 0) {
            return "";
        }
        int len = s.length();
        char[] array = new char[len * 3];
        int size = 0;
        for (int i = 0; i < len; i++) {
            char c = s.charAt(i);
            if (c == ' ') {
                array[size++] = '%';
                array[size++] = '2';
                array[size++] = '0';
            } else {
                array[size++] = c;
            }
        }
        return new String(array, 0, size);
    }

    public static String replaceSpace(String s) {
        return s.replace(" ", "%20");
    }
}