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    Java字符串、数组和二叉搜索树实例的分析

    WBOYWBOY2023-04-21 09:19:08转载28

    题目一

    Java字符串,数组及二叉搜索树实例分析

    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'&&chars[left]<='Z')){
                    tmp = chars[left];
                }else {
                    left++;
                    continue;
                }
                if(chars[right]>='a'&&chars[right]<='z'||(chars[right]>='A'&&chars[right]<='Z')){
                    chars[left] = chars[right];
                    chars[right] = tmp;
                }else {
                    right--;
                    continue;
                }
                left++;
                right--;
            }
            return new String(chars);
        }
    }

    题目二

    Java字符串,数组及二叉搜索树实例分析

    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;
     *         this.left = left;
     *         this.right = right;
     *     }
     * }
     */
    class Solution {
        public TreeNode increasingBST(TreeNode root) {
            ArrayList<Integer> list = new ArrayList<Integer>();
            method(root,list);
            TreeNode ans = new TreeNode(-1);
            TreeNode cur = ans;
            for(int i:list){
                TreeNode node = new TreeNode(i);
                cur.right = node;
                cur = cur.right;
            }
            return ans.right;
        }
        public void method(TreeNode root,List<Integer> list){
            if(root==null) return;
            method(root.left,list);
            list.add(root.val);
            method(root.right,list);
        }
    }

    题目三

    Java字符串,数组及二叉搜索树实例分析

    解法

    class Solution {
        public int[] sortArrayByParity(int[] nums) {
            int[] ans = new int[nums.length];
            int left = 0;
            int right = nums.length-1;
            for(int i : nums){
                if(i%2==0){
                    ans[left] = i;
                    left++;
                }else{
                    ans[right] = i;
                    right--;
                }
            }
            return ans;
        }
    }
     
    class Solution {
        public int[] sortArrayByParity(int[] nums) {
            int left = 0;
            int right = nums.length-1;
            while(left<=right){
                if(nums[left]%2==0){
                    left++;
                    continue;
                }
                if(nums[right]%2!=0){
                    right--;
                    continue;
                }
                if(nums[left]%2!=0&&nums[right]%2==0){
                    int tmp = nums[left];
                    nums[left] = nums[right];
                    nums[right] = tmp;
                }
            }
            return nums;
        }
    }

    题目四

    Java字符串,数组及二叉搜索树实例分析

    解法

    class Solution {
        public boolean backspaceCompare(String s, String t) {
            if(method(s).equals(method(t))) return true;
            return false;
        }
        public static String method(String s){
            int slow = 0;
            char[] chars = s.toCharArray();
            for (int i = 0; i < chars.length; i++) {
                if(chars[i]=='#'){
                    chars[i] = 0;
                    slow = i;
                    while (true){
                        if(slow-1<0) break;
                        if (chars[slow-1]!=0){
                            chars[slow-1] = 0;
                            break;
                        }
                        slow--;
                    }
                }
            }
            StringBuilder sb = new StringBuilder();
            for(char i : chars){
                if(i!=0) sb.append(i);
            }
            return sb.toString();
        }
    }

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