剑指 Offer 05. 替换空格

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：

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

限制：

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0 <= s 的长度 <= 10000
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#!/usr/bin/env python
# -*- coding: utf-8 -*-
# author：sarizzm time:2021/1/6 0006


# 暴力求解

class Solution:
    def replaceSpace(self, s):
        if s is None:
            return s

        st = str()
        for i in s:
            if i ==' ':
                st += '%20'
            else:
                st += i
        return st



print(Solution().replaceSpace('We are happy.'))


执行用时：40 ms, 在所有 Python3 提交中击败了51.35%的用户

内存消耗：14.8 MB, 在所有 Python3 提交中击败了10.11%的用户

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class Solution {
    public String replaceSpace(String s) {
        int length = s.length();
        char[] array = new char[length * 3];
        int size = 0;
        for (int i = 0; i < length; i++) {
            char c = s.charAt(i);
            if (c == ' ') {
                array[size++] = '%';
                array[size++] = '2';
                array[size++] = '0';
            } else {
                array[size++] = c;
            }
        }
        String newStr = new String(array, 0, size); // 数组技巧的操作
        return newStr;
    }
}


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