Algorithem_ReverseWords

Reverse Words in a String III

Given a string s, reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order.

Example 1:

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Input: s = "Let's take LeetCode contest"
Output: "s'teL ekat edoCteeL tsetnoc"

Example 2:

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Input: s = "God Ding"
Output: "doG gniD"

解法一

逻辑：
把字符串根据空格切割成数组，然后遍历数组，对数组中字符串调用 reversed 方法，最后在使用空格join为字符串返回

代码如下：

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class Solution {
    func reverseWords(_ s: String) -> String {
        let list = s.components(separatedBy: " ")
        var results: [String] = []
        for str in list {
            let reverseStr = String(str.reversed())
            results.append(reverseStr)
        }
        return results.joined(separator: " ")
    }
}

虽然可以得出结果，但是和没用到 TwoPointers算法，另一种解法是，使用TwoPointers，先把字符串转为字符数组，然后遍历字符数组，如果当前字符为空格，则对空格前面（两个空格之间）的元素交换位置。

代码如下：

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class Solution {
    func reverseWords(_ s: String) -> String {
        var characters = Array(s)
        var start = 0
        var end = 0
        for i in 0..<characters.count {
            let c = characters[i]
            if String(c) == " " {
                // 当前为空格，则 revese 前面的数据
                end = i - 1 
                swapList(&characters, start, end)
                start = i + 1
            }
        }
        swapList(&characters, start, characters.count-1)
        return String(characters)
    }
    
    func swapList(_ characters: inout [Character], _ start: Int, _ end: Int) {
        var mutStart = start
        var mutEnd = end
        while mutStart < mutEnd {
            characters.swapAt(mutStart, mutEnd)
            mutStart += 1
            mutEnd -= 1
        }
    }
}