算法练习，链表二分最大n个

import java.util.ArrayList;
import java.util.Collections;
import java.util.HashSet;

public class BinarySearch {
public static void main(String[] args) {
int[] a = { 11, 27, 28, 33 };
// System.out.println(findFirstRepeat("qywyer23tdd", 11));
// ListNode head = LinkedListReverse.initialList();
// LinkedListReverse.printList(insertionSortList(head));[4,5,1,6,2,7,3,8],10
GetLeastNumbers_Solution(new int[] { 4, 5, 1, 6, 2, 7, 3, 8 }, 10);
}

public static ArrayList<Integer> GetLeastNumbers_Solution(int[] input, int k) {
ArrayList<Integer> list = new ArrayList<Integer>();
int cnt = 0;
if (k > input.length || k == 0) {
return list;
}
while (cnt < k) {
list.add(input[cnt]);
cnt++;
}
Collections.sort(list);

for (int i = k; i < input.length; i++) {
int num = input[i];
// 如果超出最大，则不用管，如果没有超出最大，则需要加入并踢出最大
if (num < list.get(k - 1)) {
list.remove(k - 1);
list.add(num);
Collections.sort(list);
}
}
System.out.println(list);
return null;
}

/**
* 插入排序 head 1 -> 7 -> 2 -> 6 -> 9 1 -> 2 -> 6 -> 7 -> 9
*/
public static ListNode insertionSortList(ListNode head) {
ListNode pstart = head;
ListNode pcurr = head.next;
if (pstart == null) {
return null;
}
// //1 7 2 6 9 3 在7后面插入888
// do {
// int currVal = pstart.val;
// if (currVal == 7) {
// ListNode newNode = new ListNode(888, null);
// newNode.setNext(pstart.next);
// pstart.setNext(newNode);
// }
// } while ((pstart = pstart.next) != null);

// 1 7 2 6 9 3 在7前面插入888
do {
int currVal = pcurr.val;
if (currVal == 7) {
ListNode newNode = new ListNode(888, null);
newNode.setNext(pcurr);
pstart.setNext(newNode);
}
} while ((pcurr = pcurr.next) != null);
return head;
}

public static char findFirstRepeat(String A, int n) {
HashSet<Character> hs = new HashSet<Character>();
for (char c : A.toCharArray()) {
if (hs.contains(c)) {
return c;
} else {
hs.add(c);
}
}
return ' ';
}

public static int getPos(int[] A, int n, int val) {
int start = 0;
int end = n - 1;
int mid = (end - start) / 2;
while (start < end) {
if (A[mid] == val) {
return mid;
} else if (A[mid] > val) {
end = mid;
} else if (A[mid] < val) {
start = mid;
}
mid = start + (end - start) / 2;
}
return 0;
}
}