The sword refers to the first character position of OfferJZ34 that only appears once (JavaScript: map, indexOf+lastIndexOf)

Time limit: C/C++ 1 second, other languages ​​2 seconds Space limit: C/C++ 64M, other languages ​​128M Hot index: 486021
Knowledge points of this question: String

The title description
finds the first character that appears only once in a string (0<=string length<=10000, all composed of letters), and returns its position, if not, it returns -1 (case sensitive ). (Counting from 0)
Example 1
Input
"google" and
return value
4

Idea 1: Use map to receive each string as a key, repeat the string to get the value +1, and finally traverse the map to get the element with value === 1, and it is the first one, output
Note: forEach cannot stop the loop, but map Only this kind of loop is provided, so special treatment is needed, which means && index === -1it is a qualified element.

function FirstNotRepeatingChar(str) {
    
    
    let len = str.length
    let map = new Map()
    for (let i = 0; i < len; i++) {
    
    
        let target = map.get(str[i])
        if (target) {
    
    
            map.set(str[i], ++target)
        } else {
    
    
            map.set(str[i], 1)
        }
    }
    let index = -1
    map.forEach((val, key) => {
    
    
        if (val === 1 && index === -1) {
    
    
            index = str.indexOf(key)
        }
    })
    return index
}

Idea 2: Simply use the native subscript method of finding a string. When indexOf is a positive number, and lastIndexOf is a reverse number, if they are the same, it is a unique value.

function FirstNotRepeatingChar(str) {
    
    
    for(let i = 0 ; i < str.length;i++){
    
    
        if(str.indexOf(str[i]) == str.lastIndexOf(str[i])) return i
    }
    return -1
}