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;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
  int ans;
  int pre;
  public int minDiffInBST(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); 
  }
}

题目二

 解法

class Solution {
  public int dominantIndex(int[] nums) {
      int f = Integer.MIN_VALUE;
      int fi = 0;
      int s = Integer.MIN_VALUE;
      int si = 0;
      for(int i = 0; i<nums.length;i++){
          if(nums[i]>f){
              s = f;
              f = nums[i];
              fi = i;
          }else if(nums[i]>s){
              s = nums[i];
          }
      }
      if(nums.length==1) return 0;
      if(2*s<=f) return fi;
      return -1; 
  }
}

题目三

解法

class Solution {
  public int repeatedNTimes(int[] nums) {
      int n = nums.length/2;
      HashMap<Integer,Integer> map =new HashMap<Integer,Integer>();
      for(int key : nums){
          if(map.containsKey(key)){
              map.put(key,map.get(key)+1);
              if(map.get(key)==n){
                  return key;
              }
          }else{
              map.put(key,1);
          }
      }
      return 0;
  }
}

 题目四

 解法

class Solution {
  public boolean uniqueOccurrences(int[] arr) {
      int[] nums = new int[2000];
      for(int i =0;i<arr.length;i++){
          nums[arr[i]+1000]+=1;
      }
      HashSet<Integer> set =new HashSet<Integer>();
      for(int i =0;i<nums.length;i++){
          if(nums[i]==0) continue;
          if(!set.add(nums[i])){
              return false;
          }else{
              set.add(nums[i]);
          }
      }
      return true;
  }
}

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原文链接：https://blog.csdn.net/wai_58934/article/details/123336107