我想在数组中找到第k个最小元素,但实际上需要我的分区方法的索引。
我在这个博客上发现了这个代码,用于查找第k个最小元素: http://blog.teamleadnet.com/2012/07/quick-select-algorithm-find-kth-element.html
但这只返回值,而不是索引。
你知道如何有效地找到它的索引吗?
答案 0 :(得分:1)
最简单的方法是创建一个长度相同的indices
数组,用0
到length-1
的数字填充,当arr
数组更改时,执行与indices
数组相同的更改。最后返回indices
数组中的相应条目。您甚至不必理解原始算法来执行此操作。这是修改后的方法(我的更改标有***
):
public static int selectKthIndex(int[] arr, int k) {
if (arr == null || arr.length <= k)
throw new IllegalArgumentException();
int from = 0, to = arr.length - 1;
// ***ADDED: create and fill indices array
int[] indices = new int[arr.length];
for (int i = 0; i < indices.length; i++)
indices[i] = i;
// if from == to we reached the kth element
while (from < to) {
int r = from, w = to;
int mid = arr[(r + w) / 2];
// stop if the reader and writer meets
while (r < w) {
if (arr[r] >= mid) { // put the large values at the end
int tmp = arr[w];
arr[w] = arr[r];
arr[r] = tmp;
// *** ADDED: here's the only place where arr is changed
// change indices array in the same way
tmp = indices[w];
indices[w] = indices[r];
indices[r] = tmp;
w--;
} else { // the value is smaller than the pivot, skip
r++;
}
}
// if we stepped up (r++) we need to step one down
if (arr[r] > mid)
r--;
// the r pointer is on the end of the first k elements
if (k <= r) {
to = r;
} else {
from = r + 1;
}
}
// *** CHANGED: return indices[k] instead of arr[k]
return indices[k];
}
请注意,此方法会修改原始arr
数组。如果您不喜欢这样,请在方法的开头添加arr = arr.clone()
。