实现 indexOf() 方法

暴力解法

const fn = (long, short) => {
    let longLen = long.length
    let shortLen = short.length

    for (let i = 0; i <= longLen - shortLen; i++) {
        let j;
        for (j = 0; j < shortLen; j++) {
            if (long[i + j] !== short[j]) {
                break
            }

        }

        if (j === shortLen) return i

    }

    return -1
}

KPM

// 大概思路： 遍历long，和short不匹配，直接跳过short的长度（依赖一个进位表）
// 建立一个进位表，只和short有关
const buildMap = str => {
    const a = [...str]
    const len = a.length
    const map = {}
    a.forEach((i, index) => {
        map[i] = len - index
    })

    return map
}

// 开始操作
const fn2 = (long, short) => {
    const map = buildMap(short)

    const longLen = long.length
    const shortLen = short.length
    let i
    for (i = 0; i <= longLen - shortLen; ) {
        let j;
        for (j = 0; j < shortLen; j++) {
            if (long[i + j] !== short[j]) {
                const next = long[i + shortLen]
                const c = map[next] ? map[next] : shortLen
                i += c
                break;
            }
            
        }
        if (j === shortLen) return i
        
    }

    return -1
}