程序找到移动到前面的最小数量以排列数组中的项目并对其进行排序

Question

我正在尝试实现一种排序机制，在未排序的数组中，一次只挑选一个元素并将其放置在数组前面，需要以最少的迭代次数来实现这一点。

[3,1,3,2] //input array
[2,3,1,3]// step one
[1,2,3,3]// step tow and the result, iterations = 2

我给出的示例尝试是，这似乎不起作用 -

let sort = (inputArr) => {
    let originalArr = inputArr;
    let arrCopy = [...inputArr];
    let sortedArr = [...inputArr];
    sortedArr.sort();// need to achive here 

    let interations = 0;

    //find 2nd max number
    let secLarNum;

    Math.max(...arrCopy);

    arrCopy = removeItem(arrCopy, Math.max(...arrCopy));

    while(JSON.stringify(sortedArr) !== JSON.stringify(originalArr)) {
        let index = 0;
        secLarNum = Math.max(...arrCopy);
        arrCopy = removeItem(arrCopy, Math.max(...arrCopy));

        index = originalArr.indexOf(secLarNum);

        for(let i=0; i < originalArr.length; i ++){
            if(originalArr[index] > originalArr[i] && index !== -1){
                let currValue = originalArr[index];
                originalArr.splice(index, 1);
                originalArr.unshift(currValue);
                interations ++;
                // break;
            }
        }

    }
    console.log('Sorted array ', originalArr);
    return interations;

}


let removeItem = (array, item) => { 
    return array.filter((i) => i !== item);
}

console.log(sort([3,3,2,1,4])); // 2

Answer 1

function moveOne() {
    let idx = -1;
    let candidate = undefined;
    let max = arr[0];

    for (let i = 1; i < arr.length; i++) {
        let e = arr[i];
        if (e > max)
            max = e;

        if ((candidate == undefined || candidate < e) && e < max) {
            idx = i;
            candidate = e;
        }
    }

    if (idx == -1)
        return false;

    arr.splice(idx, 1);
    arr.unshift(candidate);
    return true;
}

let arr = [ 1, 3, 2, 5, 4, 6 ] ;
do {
    console.log(arr);
} while(moveOne());