Continuous consecutive numbers

Find continuous consecutive numbers in a string, whether they be digits or tenths

by Alan Guerrero

JavaScript

str = "1111122asdgas222*&^%121314151617bdjfjahdi234bdce56789";
consecutive = 5;
iterations = 3;
alert("It is " + consecutiveDigits(str, consecutive) + " that " + str + " contains " + consecutive + " consecutive digits.");
alert("It is " + consecutiveDigits(str, consecutive) + " that " + str + " contains " + consecutive + " consecutive tenths.");

function consecutiveDigits(str, consecutive){
    var curr,
        prev,
        count = 0;
    for(var i = 0; i < str.length; ++i) {
        curr = parseInt(str.split('')[i]);
        if(isNumeric(curr)) {
            if(count === 0){
                ++count;
            }
            else if(prev + 1 === curr){
                ++count;
                if(count === consecutive){
                    return true;
                }
            }
            prev = curr;
        }
    }
    return false;
}

function consecutiveTenths(str, consecutive, iterations){
    var curr,
        prev,
        curr_tenth = 0,
        prev_tenth = 0,
        count = 0,
        count_tenth = 0;
    
    for(var i = 0; i < str.length; ++i) {
        curr = parseInt(str.split('')[i]);
        if(isNumeric(curr)) {
            ++count;
            if(count === iterations){
                curr_digit = (prev * 10) + curr;
                alert(count_digit + " " + curr_digit + " " + prev_tenth);
                if(count_digit === 0){
                    ++count_digit;
                }
                else if(curr_tenth === (prev_tenth + 1)){
                    ++count_digit;
                    if(count_digit === consecutive){
                        return true;
                    }
                }
                prev_digit = curr_digit;
                count = 0;
            }
            else {
                prev = curr;
            }
        }
        else {
            count = 0;
        }
    }
}


function isNumeric(n) {
    return !isNaN(parseFloat(n)) && isFinite(n);
}