string - Swift 快速低级 String lastIndexOf

Question

我需要一个尽可能快的 lastIndexOf 实现。我发现 String advance 函数非常慢。我尝试使用 c 函数 strrchr，并尝试将字符串复制到 NSData 并使用指针，但我无法获得正确的语法。我的字符串将始终包含 1 个字节的字符和我正在搜索的字符串“|”也总是 1 个字节。

任何使用 advance 的实现都会太慢，但这是我能找到的最快的例子:

func indexOf(target: String, startIndex: Int) -> Int
{
    var startRange = advance(self.startIndex, startIndex)

    var range = self.rangeOfString(target, options: NSStringCompareOptions.LiteralSearch, range: Range<String.Index>(start: startRange, end: self.endIndex))

    if let range = range {
        return distance(self.startIndex, range.startIndex)
    } else {
        return -1
    }
}

func lastIndexOf(target: String) -> Int
{

    var index = -1
    var stepIndex = self.indexOf(target)
    while stepIndex > -1
    {
        index = stepIndex
        if stepIndex + target.length < self.length
        {
            stepIndex = indexOf(target, startIndex: stepIndex + target.length)
        }
        else
        {
            stepIndex = -1
        }
    }
    return index
}

这是我需要解析的字符串示例。 var str:String = "4|0|66|5|0|3259744|6352141|1|3259744|WSMxt208L54yZ5irtHC3|Mc02|efland,nc|36.027992|-79.2212834|0|4|6|0|3259744|6352141|46 |14|1|0|7|7|3259744|6352141|4|1|0|8|8|3259744|6352141|4|0|22|9|0|3259744|6352141|2|3|房间1|2 |72|86330534|1|0|10|9|3259744|6352141|4|1|0|11|10|3259744|6352141|4|1|0|12|11|3259744|6352141|4|1|0 |13|12|3259744|6352141|4|0|4|14|0|3259744|6352141|46|24|0|5|15|0|3259744|6352141|46|654|0|66|0|0 |3259744|6352141|1|3259744|WSMxt208L54yZ5irtHC3|Mc02|efland,nc|36.027992|-79.2212834|0|4|16|0|3259744|6352141|46|4sageReceived:4|0|66|9|74|425 6352141|1|3259744|WSMxt208L54yZ5irtHC3|Mc02|efland,nc|36.027992|-79.2212834|0|4|6|0|3259744|6352141|46|14|1|0|7|7|3259744|6352141|4|1 |0|8|8|3259744|6352141|4|0|22|9|0|3259744|6352141|2|3|房间1|2|72|86330534|1|0|10|9|3259744|6352141|4 |1|0|11|10|3259744|6352141|4|1|0|12|11|3259744|6352141|4|1|0|13|12|3259744|6352141|4|0|4|14|0 |3259744|6352141|46|24|0|5|15|0|3259744|6352141|46|654|0|66|0|0|3259744|6352141|1|3259744|WSMxt208L54y Z5irtHC3|Mc02|efland,nc|36.027992|-79.2212834|0|4|16|0|3259744|6352141|46|4352141|1|3259744|WSMxt208L54yZ5irtHC3|Mc02|efland,nc|36.027922|243.21|243.79 6|0|3259744|6352141|46|14|1|0|7|7|3259744|6352141|4|1|0|8|8|3259744|6352141|4|0|22|9|0|3259744| 6352141|2|3|房间1|2|72|86330534|1|0|10|9|3259744|6352141|4|1|0|11|10|3259744|6352141|4|1|0|12|11| 3259744|6352141|4|1|0|13|12|3259744|6352141|4|0|4|14|0|3259744|6352141|46|24|0|5|15|0|3259744|6352141|46| 654|0|66|0|0|3259744|6352141|1|3259744|WSMxt208L54yZ5irtHC3|Mc02|efland,nc|36.027992|-79.2212834|0|4|16|0|3259744|6352141|46|4TCPLivedener:.onRe |0|66|5|0|3259744|6352141|1|3259744|WSMxt208L54yZ5irtHC3|Mc02|efland,nc|36.027992|-79.2212834|0|4|6|0|3259744|6352141|46|14|1|0| 7|7|3259744|6352141|4|1|0|8|8|3259744|6352141|4|0|22|9|0|3259744|6352141|2|3|房间1|2|72|86330534|1| 0|10|9|3259744|6352141|4|1|0|11|10|3259744|6352141|4|1|0|12|11|3259744|6352141|4|1|0|13|12|3259744| 6352141|4|0|4|14|0|3259744|6352141|46|24|0|5|15|0|3259744|6352141|46|654 |0|66|0|0|3259744|6352141|1|3259744|WSMxt208L54yZ5irtHC3|Mc02|efland,nc|36.027992|-79.2212834|0|4|16|0|3259744|6352141|46|4preParse4|0|66|5|0|3259744|6352141|1|3259744|WSMxt208L54yZ5irtHC3|Mc02|efland,nc|36.027992|-79.221283"

Answer 1

这是一个 Swift 2.0 的答案

func lastIndexOf(s: String) -> Int? {
    if let r: Range<Index> = self.rangeOfString(s, options: .BackwardsSearch) {
        return self.startIndex.distanceTo(r.startIndex)
    }

     return Optional<Int>()
}

测试

func testStringLastIndexOf() {
    let lastIndex = "0|2|45|7|9".lastIndexOf("|")

    XCTAssertEqual(lastIndex, 8)
}

func testStringLastIndexOfNotFound() {
    let lastIndex = "0123456789".lastIndexOf("|")

    XCTAssertEqual(lastIndex, nil);
}