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'&&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

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

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

 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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原文链接：https://blog.csdn.net/wai_58934/article/details/123306957

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