题解|高效率求出乱序数组中的第k小元素

高效率求出乱序数组中的第k小元素

public class 最快效率求出乱序数组中的第k小元素 {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		int[] arr = { 9, 8, 4, 3215, 1, 54, 9, 3 };
		System.out.println(f(arr, 2));
	}

	private static int f(int[] arr, int k) {
		int begin = 0;
		int end = arr.length - 1;
		while (true) {
			int indexThVal = partition(arr, begin, end);// 分区并返回数组主元的索引,[begin,end]数组中第(indexThVal + 1)小的位置是(indexThVal).
			if (k < indexThVal + 1) {
				end = indexThVal - 1;
			} else if (k > indexThVal + 1) {
				begin = indexThVal + 1;
			} else {
				return arr[k - 1];
			}
		}
	}

	private static int partition(int[] arr, int begin, int end) {
		int pivot = arr[begin];
		int left = begin + 1;
		int right = end;
		while (left <= right) {
			while (left <= right && arr[left] <= pivot) {
				left++;
			}
			while (left <= right && arr[right] > pivot) {
				right--;
			}
			if (left < right) {
				swap(arr, left, right);
			}
		}
		swap(arr, begin, right);
		return right;
	}

	private static void swap(int[] arr, int i, int j) {
		int temp = arr[i];
		arr[i] = arr[j];
		arr[j] = temp;
	}
}

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转载自blog.csdn.net/a739260008/article/details/86562307