Swift.替换空格

题目：请实现一个函数，把字符串中的每个空格替换成"%20"。例如输入"We are happy."，则输出"We%20are%20happy."。

class Solution {
    func replace(_ s: String) -> String {
        
        guard s.count > 0 else {
            return s
        }
        
        var nullNum = 0
        for c in s {
            if c == " " {
                nullNum += 1
            }
        }
        
        let len = s.count + nullNum * 2
        var list = Array<Character>(repeating: "x", count: len)
        
        var listIndex = len - 1
        for (_,c) in s.enumerated().reversed() {
            if c == " " {
                list[listIndex] = "0"
                listIndex -= 1
                list[listIndex] = "2"
                listIndex -= 1
                list[listIndex] = "%"
                listIndex -= 1
            }else {
                list[listIndex] = c
                listIndex -= 1
            }
        }
        
        return String(list)
    }
}

解题思路：如果从前往后处理，则空格后面所有的字符都后移若干位。所以从后向前处理会有更优解。这里直接新建了一个数组模拟操作的。

github地址：https://github.com/cubegao/LeetCode