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如何找到在javascript中按降序对数字数组进行排序所需的最小交换次数

[英]How to find the Minimum number of swaps required for sorting an array of numbers in descending order in javascript

I'm trying to get my code to do this:我试图让我的代码做到这一点:

Original array = [1,2,3,4] swap once-> [4,2,3,1] swap again->[4,3,2,1]原始数组 = [1,2,3,4] 交换一次-> [4,2,3,1] 再次交换->[4,3,2,1]

Therefore result is 2因此结果是 2

But it's not working.但它不起作用。 Here's what I have so far:这是我到目前为止所拥有的:

 function check(arr){ var sarr = []; var cnt = 0; var arrL = arr.length; // Create a second copy of the array for reference var arrCopy = [...arr]; for(let i=0; i<arrL;i++){ var maxV = Math.max(...arr); sarr.push(maxV); let pos = arr.indexOf(maxV); // Remove the found number arr.splice(pos,1); // Check if the index of the number in the new array is same with the copy, if not then there was a swap let ai =arrCopy.indexOf(maxV); let si =sarr.indexOf(maxV); if (ai !== si && (i+1)!=arrL && pos !== 0){ cnt++; }; } console.log(cnt); } check([1, 2, 3, 4, 5, 6]);//Result should be 3 check([6,5,4,3,2,1]); //result should be 0 check([1,2,3,4]); //result should be 2 check([1,3,2,5,4,6]); //result should be 3 check([1,2,10,4,5,6,7,8,9,3,12,11]);//result should be 6 check([ 49, 37, 9, 19, 27, 3, 25, 11, 53,  42, 57, 50, 55,  56, 38, 48, 6, 33, 28, 8, 20, 31, 51, 14, 23, 4, 58, 52, 36, 22, 41, 47, 39, 2, 7, 13, 45, 1, 44, 32, 10, 15, 21, 30, 17,  60, 29, 5, 59, 12, 40, 24, 54, 46, 26, 43, 35, 34, 18, 16]);//result should be 54

Can someone please let me know what I'm doing wrong?有人可以让我知道我做错了什么吗?

I would start with a copy of the array in descending order for getting the right index of the items.我将从按降序排列的数组副本开始,以获得正确的项目索引。

For practical reasons, (or just a shorter conception of the loop with including check and decrement), I loop from the end of the array.出于实际原因,(或者只是包含检查和递减的循环的较短概念),我从数组的末尾开始循环。

Then I check the value of array and reversed at the dame index and go on with the iteration.然后我检查array的值并在夫人索引处reversed并继续迭代。

If not the same value, the items at the wanted position i and the actual position p are swapped and the count incremented.如果不是相同的值,则交换所需位置i和实际位置p项目并增加计数。

At the end the count is returned.最后返回计数。

 function check(array) { var reversed = array.slice().sort((a, b) => b - a), count = 0, i = array.length, p; while (i--) { if (array[i] === reversed[i]) continue; p = array.indexOf(reversed[i]); [array[i], array[p]] = [array[p], array[i]]; count++; } console.log(...array); return count; } console.log(check([1, 2, 3, 4, 5, 6])); // 3 console.log(check([6, 5, 4, 3, 2, 1])); // 0 console.log(check([1, 2, 3, 4])); // 2 console.log(check([1, 3, 2, 5, 4, 6])); // 3 console.log(check([1, 2, 10, 4, 5, 6, 7, 8, 9, 3, 12, 11])); // 6 console.log(check([ 49, 37, 9, 19, 27, 3, 25, 11, 53,  42, 57, 50, 55,  56, 38, 48, 6, 33, 28, 8, 20, 31, 51, 14, 23, 4, 58, 52, 36, 22, 41, 47, 39, 2, 7, 13, 45, 1, 44, 32, 10, 15, 21, 30, 17,  60, 29, 5, 59, 12, 40, 24, 54, 46, 26, 43, 35, 34, 18, 16])); // 54
 .as-console-wrapper { max-height: 100% !important; top: 0; }

function minimumSwaps(arr) {
  var count = 0;
  arr.sort((a, b) => {
    if (a < b) {
      count++;
    }
  });
  return count;
}

console.log(minimumSwaps([1, 2, 3, 4, 7, 6, 5]));

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