剑指Offer之Java算法习题精讲字符串与二叉搜索树

题目一

解法

class Solution {
    public boolean repeatedSubstringPattern(String a) {
        for (int i = 1; i <=a.length()/2 ; i++) {
            String s = a.substring(0, i);
            StringBuffer sb = new StringBuffer();
            while (sb.length()<a.length()){
                sb.append(s);
            }
            if(sb.toString().equals(a)){
                return true;
            }
        }
        return false;
    }
}

题目二

解法

class Solution {
    public boolean detectCapitalUse(String word) {
        if(word.toLowerCase().equals(word)||word.toUpperCase().equals(word)||word.substring(1, word.length()).toLowerCase().equals(word.substring(1, word.length()))) return true;
        return false;
    }
}

题目三

解法

二叉搜索树有个性质为二叉搜索树中序遍历得到的值序列是递增有序的

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    int ans;
    int pre;
    public int getMinimumDifference(TreeNode root) {
        ans = Integer.MAX_VALUE;
        pre = -1;
        method(root);
        return ans;
    }
    public void method(TreeNode root){
        if(root==null){
            return;
        }
        method(root.left);
        if(pre==-1){
            pre = root.val;
        }else{
            ans = Math.min(ans, root.val - pre);
            pre = root.val;
        }
        method(root.right);
    }
}

题目四

解法

class Solution {
    public String reverseStr(String a, int k) {
        int con = 1;
        StringBuffer sb = new StringBuffer();
        while (con*k<=a.length()){
            String substring = a.substring((con - 1) * k, con * k);
            if(con%2==0){
                sb.append(substring);
                con++;
            }else {
                for (int i1 = substring.length()-1; i1 >=0 ; i1--) {
                    sb.append(substring.charAt(i1));
                }
                con++;
            }
        }
        if((con-1)*k<a.length()){
            String s = a.substring((con-1) * k, a.length());
            if(con%2!=0){
                for (int i1 = s.length()-1; i1>=0; i1--) {
                    sb.append(s.charAt(i1));
                }
            }else {
                sb.append(s);
            }
        }
        return sb.toString();
    }
}

到此这篇关于剑指Offer之Java算法习题精讲字符串与二叉搜索树的文章就介绍到这了,更多相关Java 二叉搜索树内容请搜索我们以前的文章或继续浏览下面的相关文章希望大家以后多多支持我们!

(0)

相关推荐

  • 剑指Offer之Java算法习题精讲二叉搜索树与数组查找

    题目一  解法 /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; *

  • 剑指Offer之Java算法习题精讲二叉树专项解析

    题目一 解法 /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * t

  • 剑指Offer之Java算法习题精讲数组与字符和等差数列

    题目一  解法 class Solution { public int[] relativeSortArray(int[] arr1, int[] arr2) { int[] arr = new int[1001]; int[] ans = new int[arr1.length]; int index = 0; for(int i =0;i<arr1.length;i++){ arr[arr1[i]]+=1; } for(int i = 0;i<arr2.length;i++){ while

  • 剑指Offer之Java算法习题精讲数组与列表的查找及字符串转换

    题目一 解法 class Solution { public String toLowerCase(String s) { StringBuilder sb = new StringBuilder(); for(int i = 0;i<s.length();i++){ char ch = s.charAt(i); if('A'<=ch&&ch<='Z'){ ch = (char)(ch+32); } sb.append(ch); } return sb.toString(

  • 剑指Offer之Java算法习题精讲数组与字符串

    题目一  解法 class Solution { public int findLengthOfLCIS(int[] nums) { if(nums.length==1) return 1; int fast = 1; int tmp = 1; int max = Integer.MIN_VALUE; while(fast<nums.length){ if(nums[fast]>nums[fast-1]){ tmp++; max = Math.max(max,tmp); }else{ max

  • 剑指Offer之Java算法习题精讲字符串操作与数组及二叉搜索树

    题目一  解法 class Solution { public String reverseOnlyLetters(String s) { char[] chars = s.toCharArray(); int left = 0; int right = chars.length-1; while(left<=right){ char tmp = 0; if(chars[left]>='a'&&chars[left]<='z'||(chars[left]>='A'&

  • 剑指Offer之Java算法习题精讲排列与N叉树

    题目一  解法 class Solution { LinkedList<List<Integer>> ans = new LinkedList<List<Integer>>(); public List<List<Integer>> permute(int[] nums) { LinkedList<Integer> list = new LinkedList<Integer>(); boolean[] bo =

  • 剑指Offer之Java算法习题精讲求和篇

    题目一   解法 /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode() {} * ListNode(int val) { this.val = val; } * ListNode(int val, ListNode next) { this.val = val; this.next = next; } * } */ class Solu

  • 剑指Offer之Java算法习题精讲N叉树的遍历及数组与字符串

    题目一 解法 /* // Definition for a Node. class Node { public int val; public List<Node> children; public Node() {} public Node(int _val) { val = _val; } public Node(int _val, List<Node> _children) { val = _val; children = _children; } }; */ class S

  • 剑指Offer之Java算法习题精讲字符串与二叉搜索树

    题目一 解法 class Solution { public boolean repeatedSubstringPattern(String a) { for (int i = 1; i <=a.length()/2 ; i++) { String s = a.substring(0, i); StringBuffer sb = new StringBuffer(); while (sb.length()<a.length()){ sb.append(s); } if(sb.toString(

  • 剑指Offer之Java算法习题精讲链表专题篇

    题目一  解法 /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode() {} * ListNode(int val) { this.val = val; } * ListNode(int val, ListNode next) { this.val = val; this.next = next; } * } */ class Solut

  • 剑指Offer之Java算法习题精讲二叉树与斐波那契函数

    题目一 解法 class Solution { public int fib(int n) { int[] arr = new int[31]; arr[0] = 0; arr[1] = 1; for(int i = 2;i<=n;i++){ arr[i] = arr[i-2]+arr[i-1]; } return arr[n]; } } 题目二  解法 /** * Definition for a binary tree node. * public class TreeNode { * in

  • 剑指Offer之Java算法习题精讲二叉树与N叉树

    题目一  解法 /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; *

  • 剑指Offer之Java算法习题精讲二叉树专题篇下

    题目一  解法 /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; *

  • 剑指Offer之Java算法习题精讲二叉树专题篇上

    来和二叉树玩耍吧~ 题目一 解法 /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val

随机推荐