除了一些nameming混亂,如pivotelement
vs pivot
和Math.round
vs Math.floor
,則需要解決例如[1, 1, 1]
這個始終返回left = []
和right = [1, 1, 1]
的問題,該問題會無限次地調用quicksort([1, 1, 1])
。
要解決此問題,您需要檢查空的left
和right
的每個元素,如果它等於隨機數據透視表。然後返回right
,而不必再撥打quicksort
。
function quicksort(array) {
var randomPlace = Math.floor(Math.random() * array.length),
pivot = array[randomPlace],
left = [],
right = [],
i;
for (i = 0; i < array.length; i++) {
(array[i] < pivot ? left : right).push(array[i]);
}
console.log(pivot, JSON.stringify(array), JSON.stringify(left), JSON.stringify(right));
// prevent looping forever
if (!left.length && right.every(function (v) { return v === pivot; })) {
return right;
}
if (left.length <= 1 && right.length <= 1) {
return left.concat(right);
}
if (left.length <= 1) {
return left.concat(quicksort(right));
}
if (right.length <= 1) {
return quicksort(left).concat(right);
}
return quicksort(left).concat(quicksort(right));
}
console.log(quicksort([2, 7, 4, 8, 3, 11, 49, 20, 10, 1, 1, 1]));
.as-console-wrapper { max-height: 100% !important; top: 0; }
另一種解決方案是將陣列分成三個陣列,一個用於更小的值,一個用於相等的值和一個更大的值。然後只對較小和較大的數組進行排序。
function quicksort(array) {
var randomPlace = Math.floor(Math.random() * array.length),
pivotValue = array[randomPlace],
left = [],
pivot = [],
right = [],
i;
for (i = 0; i < array.length; i++) {
if (array[i] === pivotValue) {
pivot.push(array[i]);
continue;
}
(array[i] < pivotValue ? left : right).push(array[i]);
}
console.log(pivotValue, JSON.stringify(array), JSON.stringify(left), JSON.stringify(pivot), JSON.stringify(right));
if (left.length <= 1 && right.length <= 1) {
return left.concat(pivot, right);
}
if (left.length <= 1) {
return left.concat(pivot, quicksort(right));
}
if (right.length <= 1) {
return quicksort(left).concat(pivot, right);
}
return quicksort(left).concat(pivot, quicksort(right));
}
console.log(quicksort([2, 7, 4, 8, 3, 11, 49, 20, 10, 1, 1, 1]));
.as-console-wrapper { max-height: 100% !important; top: 0; }
的可能的複製[JavaScript的快速排序(http://stackoverflow.com/questions/5185864/javascript-quicksort) – Liam