leetcode 665

Posted 2021-01-25 热之雪

665. Non-decreasing Array

Input: [4,2,3]
Output: True
Explanation: You could modify the first 4 to 1 to get a non-decreasing array.递增

思路：贪心思想，找异常值，存在两个以上则返回false。如果当前的值比前一的值小，并且也比前两的值小，此时只需更改当前值，而不更改前两个值。

更改规则是用前一的值代替当前值。

两种情况

例如： 2   2   1  -》  2  2  2

            0   2   1  -》  0  1  1

bool checkPossibility(vector<int>& nums) {
        int cnt = 0;                                                                    //如果存在两个以上的异常值则直接返回false
        for(int i = 1; i < nums.size() && cnt<=1 ; i++){
            if(nums[i-1] > nums[i]){                                                    //存在异常值
                cnt++;
                if(i-2<0 || nums[i-2] <= nums[i])nums[i-1] = nums[i];                    //i-2处理第一个边界值，这种||技巧经常用到
                else nums[i] = nums[i-1];                                                //have to modify nums[i]
            }
        }
        return cnt<=1;
    } 

 

669. Trim a Binary Search Tree

Input: 
    3
   / \\
  0   4
   \\
    2
   /
  1

  L = 1
  R = 3

Output: 
      3
     / 
   2   
  /
 1

 

The code works as recursion.

If the root value in the range [L, R]
      we need return the root, but trim its left and right subtree;
else if the root value < L
      because of binary search tree property, the root and the left subtree are not in range;
      we need return trimmed right subtree.
else
      similarly we need return trimmed left subtree.

Without freeing memory

class Solution {
public:
    TreeNode* trimBST(TreeNode* root, int L, int R) {
        if (root == NULL) return NULL;
        if (root->val < L) return trimBST(root->right, L, R);
        if (root->val > R) return trimBST(root->left, L, R);
        root->left = trimBST(root->left, L, R);
        root->right = trimBST(root->right, L, R);
        return root;
    }
};

 

 总结：树的遍历bfs，一般结构：

 当前root

////////////////////////////

中间逻辑，比如深一层的遍历

///////////////////////////

 对当前root进行操作，例如左右边子树的赋值操作