• 中信卡笔试代码


    LeetCode 69. x 的平方根

    实现 int sqrt(int x) 函数。

    计算并返回 x 的平方根,其中 x 是非负整数。

    由于返回类型是整数,结果只保留整数的部分,小数部分将被舍去。

    示例 1:

    输入: 4
    输出: 2
    
    示例 2:
    
    输入: 8
    输出: 2
    说明: 8 的平方根是 2.82842..., 
    由于返回类型是整数,小数部分将被舍去。
    

    Coding:

    public class Sqrt {

    package com.leetcode.two;
    
    import java.util.Scanner;
    
    /**
     * @Auther: Jibny Zhan
     * @Date: 2019/12/3 20:56
     * @Description:
     */
    public class Sqrt {
    
        public static void main(String[] args) {
            Scanner sc = new Scanner(System.in);
            int input = sc.nextInt();
            System.out.println(mySqrt2(input));
        }
    
        //暴力
        private static int mySqrt(int x) {
            if (x <= 1) return x;
            for (int i = 2; i <= x; i++)
                if (i > x / i)
                    return i - 1;
            return -1;
        }
    
        //二分法
        private static int mySqrt1(int x) {
            if (x <= 1) return x;
            int left = 0, right = x;
            while (left <= right) {
                int mid = (right + left) >>> 1;
                if (mid > x / mid)
                    right = mid - 1;
                else
                    left = mid + 1;
            }
            return right;
        }
    
        //牛顿迭代法
        public static int mySqrt2(int x) {
            int res = x;
            while (res > x / res) {
                res = (res + x / res) / 2;
            }
            return res;
        }
    
    }
    
    

    }

    LeetCode 109. 有序链表转换二叉搜索树

    给定一个单链表,其中的元素按升序排序,将其转换为高度平衡的二叉搜索树。

    本题中,一个高度平衡二叉树是指一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过 1。

    示例:

    给定的有序链表: [-10, -3, 0, 5, 9],
    
    一个可能的答案是:[0, -3, 9, -10, null, 5], 
    它可以表示下面这个高度平衡二叉搜索树:
         0
        / 
      -3   9
      /   /
    -10  5
    

    三种解法

    • 二叉树中序遍历数组有序
    • 链表快慢指针
    • 数组二分递归

    Coding:

    package com.leetcode;
    
    
    import java.util.ArrayList;
    import java.util.List;
    
    /**
     * @Auther: Jibny Zhan
     * @Date: 2019/12/3 21:18
     * @Description:
     */
    public class SortedListToBST {
    
        private static ListNode pNode;
    
        public static void main(String[] args) {
            ListNode head = new ListNode(0);
            initListNode(head);
    //      TreeNode treeNode = sortedArrayToBST(head);
    //      TreeNode treeNode = sortedListToBST(head);
            TreeNode treeNode = sortedTreeToBST(head);
            prePrintTree(treeNode);
            System.out.println();
            midPrintTree(treeNode);
        }
    
        //中序遍历二叉树得到的就是从小到大的有序数组
        private static TreeNode sortedTreeToBST(ListNode head) {
            int size = findSize(head);
            pNode = head;
            return sortedTreeToBST(0, size - 1);
        }
    
        private static int findSize(ListNode head) {
            int size = 0;
            while (head != null) {
                head = head.next;
                size++;
            }
            return size;
        }
    
        private static TreeNode sortedTreeToBST(int start, int end) {
            if (start > end) return null;
    
            int mid = start + ((end - start) >> 1);
            //划分左右子树
            TreeNode left = sortedTreeToBST(start, mid - 1);
    
            TreeNode treeNode = new TreeNode(pNode.val);
            treeNode.left = left;
    
            pNode = pNode.next;
    
            TreeNode right = sortedTreeToBST(mid+1,end);
            treeNode.right = right;
    
            return treeNode;
        }
    
    
    
        //快慢指针
        private static TreeNode sortedListToBST(ListNode head) {
            return sortedListToBST(head, null);
        }
    
    
        private static TreeNode sortedListToBST(ListNode head, ListNode tail) {
            if (head == null || head == tail)
                return null;
            ListNode slow = head;
            ListNode fast = head;
    
            while (fast.next != tail && fast.next.next != tail) {
                fast = fast.next.next;
                slow = slow.next;
            }
    
            TreeNode treeNode = new TreeNode(slow.val);//根节点
            treeNode.left = sortedListToBST(head, slow);
            treeNode.right = sortedListToBST(slow.next, tail);
    
            return treeNode;
        }
    
    
        //转成数组,数组二分,递归构建中心节点
        private static TreeNode sortedArrayToBST(ListNode head) {
            List<Integer> arrayList = new ArrayList<>();
            ListNode p = head;
            while (p != null) {
                arrayList.add(p.val);
                p = p.next;
            }
            return sortedArrayToBST(arrayList, 0, arrayList.size());
        }
    
        private static TreeNode sortedArrayToBST(List<Integer> arr, int left, int right) {
            if (left == right) return null;
            int mid = (left + right) >>> 1;
            TreeNode treeNode = new TreeNode(arr.get(mid));
            treeNode.left = sortedArrayToBST(arr, left, mid);
            treeNode.right = sortedArrayToBST(arr, mid + 1, right);
            return treeNode;
        }
    
    
        //初始化
    
        public static class ListNode {
            int val;
            ListNode next;
    
            ListNode(int x) {
                val = x;
            }
        }
    
        public static class TreeNode {
            int val;
            TreeNode left;
            TreeNode right;
    
            TreeNode(int x) {
                val = x;
            }
        }
    
        //前序打印
        public static void prePrintTree(TreeNode binTree) {
            if (binTree != null) {
                System.out.println(binTree.val);
                prePrintTree(binTree.left);
                prePrintTree(binTree.right);
            }
        }
    
        //中序打印
        public static void midPrintTree(TreeNode binTree) {
            if (binTree != null) {
                prePrintTree(binTree.left);
                System.out.println(binTree.val);
                prePrintTree(binTree.right);
            }
        }
    
        public static ListNode initListNode(ListNode head) {
            ListNode p = head;
            p.val = -10;
            p.next = new ListNode(-3);
            p = p.next;
            p.next = new ListNode(0);
            p = p.next;
            p.next = new ListNode(5);
            p = p.next;
            p.next = new ListNode(9);
            p = p.next;
            p.next = null;
            return head;
        }
    
    }
    
    
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  • 原文地址:https://www.cnblogs.com/binjz/p/12501325.html
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