LeetCode|LeetCode 每日一题 [40] 替换空格 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 【LeetCode|LeetCode 每日一题 [40] 替换空格】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"); } }