[英]Find the Kth smallest element. looping
我嘗試使用我的代碼找到第k個最小元素,但無法修復我的代碼中的錯誤。 當它嘗試使用pivot = 0對[0,0,2]進行分區時,它會循環。
import java.util.Arrays;
public class OrderStat {
public static void main(String[] args) {
int[] uA = {13, 32, 28, 17, 2, 0, 14, 34, 35, 0};
System.out.println("Initial array: " + Arrays.toString(uA));
int kth = 3; // We will try to find 3rd smallest element(or 2nd if we will count from 0).
int result = getKthSmallestElement(uA, 0, uA.length - 1, kth - 1);
System.out.println(String.format("The %d smallest element is %d", kth, result));
System.out.println("-------------------------------------");
Arrays.sort(uA);
System.out.println("Sorted array for check: " + Arrays.toString(uA));
}
private static int getKthSmallestElement(int[] uA, int start, int end, int kth) {
int l = start;
int r = end;
int pivot = uA[start];
System.out.println("===================");
System.out.println(String.format("start=%d end=%d", start, end));
System.out.println("pivot = " + pivot);
//ERROR HERE: When we will work with [0, 0, 2] part of array with pivot = 0 it will give us infinite loop;
while (l < r) {
while (uA[l] < pivot) {
l++;
}
while (uA[r] > pivot) {
r--;
}
if (l < r) {
int tmp = uA[l];
uA[l] = uA[r];
uA[r] = tmp;
}
}
System.out.println("After partitioning: " + Arrays.toString(uA) + "\n");
if (l < kth)
return getKthSmallestElement(uA, l + 1, end, kth);
else if (l > kth)
return getKthSmallestElement(uA, start, l - 1, kth);
return uA[l];
}
}
請解釋一下,如何解決這個問題。
交換后
if (l < r) {
int tmp = uA[l];
uA[l] = uA[r];
uA[r] = tmp;
}
你需要將l
和r
(或至少其中一個,以取得任何進展)移動到下一個位置( ++l; --r;
)。 否則,如果兩個值都等於pivot,則無限循環。
可以在快速排序中使用的正確分區
// make sure to call it only with valid indices, 0 <= start <= end < array.length
private int partition(int[] array, int start, int end) {
// trivial case, single element array - garbage if end < start
if(end <= start) return start;
int pivot = array[start]; // not a good choice of pivot in general, but meh
int left = start+1, right = end;
while(left < right) {
// move left index to first entry larger than pivot or right
while(left < right && array[left] <= pivot) ++left;
// move right index to last entry not larger than pivot or left
while(right > left && array[right] > pivot) --right;
// Now, either
// left == right, or
// left < right and array[right] <= pivot < array[left]
if (left < right) {
int tmp = array[left];
array[left] = array[right];
array[right] = tmp;
// move on
++left;
--right;
}
}
// Now left >= right.
// If left == right, we don't know whether array[left] is larger than the pivot or not,
// but array[left-1] certainly is not larger than the pivot.
// If left > right, we just swapped and incremented before exiting the loop,
// so then left == right+1 and array[right] <= pivot < array[left].
if (left > right || array[left] > pivot) {
--left;
}
// Now array[i] <= pivot for start <= i <= left, and array[j] > pivot for left < j <= end
// swap pivot in its proper place in the sorted array
array[start] = array[left];
array[left] = pivot;
// return pivot position
return left;
}
然后你可以在數組中找到第k個最小的元素
int findKthSmallest(int array, int k) {
if (k < 1) throw new IllegalArgumentException("k must be positive");
if (array.length < k) throw new IllegalArgumentException("Array too short");
int left = 0, right = array.length-1, p;
--k; // 0-based indices
while(true) {
p = partition(array, left, right);
if (p == k) return array[p];
if (p < k) {
left = p+1;
k -= left;
} else {
right = p-1;
}
}
// dummy return, never reached
return 0;
}
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