1. LeetCode刷题笔记(2022-05)
  2. LeetCode刷题笔记(2022-06)
  3. LeetCode刷题笔记(2022-07,08)
  4. LeetCode刷题笔记(2022-09)
  5. LeetCode刷题笔记(2022-11)

[461] 汉明距离

Time: 2022-06-02

Difficulty: Easy (81.64%)

Tags: bit-manipulation

两个整数之间的汉明距离指的是这两个数字对应二进制位不同的位置的数目。

非常简单

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
class Solution {
public:
  /**
   * @brief
   * 149/149 cases passed (0 ms)
   * Your runtime beats 100 % of cpp submissions
   * Your memory usage beats 65.83 % of cpp submissions (5.8 MB)
   * @param x
   * @param y
   * @return int
   */
  int hammingDistance(int x, int y) {
    int res = x % 2 != y % 2;
    for (int i = 0; i < 31; i++)
      res += (x >>= 1) % 2 != (y >>= 1) % 2;
    return res;
  }
};

[494] 目标和

Time: 2022-06-02

Difficulty: Medium (49.03%)

Tags: dynamic-programming | depth-first-search

自己没做出来，运行超时了

首先贴一下自己的方法

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
class MySolution {
private:
  int hammingDistance(int x, int y) {
    int res = x % 2 != y % 2;
    for (int i = 0; i < 31; i++)
      res += (x >>= 1) % 2 != (y >>= 1) % 2;
    return res;
  }

public:
  /**
   * @brief
   * 超时
   * @param nums
   * @param target
   * @return int
   */
  int findTargetSumWays(vector<int> &nums, int target) {
    auto possibilities = 1;
    long long max = 0;
    for (auto num : nums) {
      max += num;
      possibilities <<= 1;
    }
    vector<long long> dp(possibilities);
    dp[possibilities - 1] = max;
    int count = max == target;
    for (int i = possibilities - 2; i >= 0; i--) {
      for (int j = i + 1; j < possibilities; j++) {
        if (hammingDistance(i, j) == 1) {
          auto sub = j - i, tmp = 0;
          while (sub > 0) {
            sub >>= 1;
            tmp += 1;
          }
          dp[i] = dp[j] - 2 * (nums[nums.size() - tmp]);
          count += dp[i] == target;
          break;
        }
      }
    }
    return count;
  }
};

然后是正解

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
class Solution {
public:
  /**
   * @brief
   * 139/139 cases passed (4 ms)
   * Your runtime beats 97.46 % of cpp submissions
   * Your memory usage beats 39.23 % of cpp submissions (9.1 MB)
   * sum(P) 前面符号为+的集合；sum(N) 前面符号为减号的集合
   * 所以题目可以转化为
   * sum(P) - sum(N) = target
   * => sum(nums) + sum(P) - sum(N) = target + sum(nums)
   * => 2 * sum(P) = target + sum(nums)
   * => sum(P) = (target + sum(nums)) / 2
   * 因此题目转化为01背包，也就是能组合成容量为sum(P)的方式有多少种
   * @param nums
   * @param target
   * @return int
   */
  int findTargetSumWays(vector<int> &nums, int target) {
    int sum = 0;
    for (auto num : nums)
      sum += num;
    if (sum < target || -sum > target || (sum + target) % 2 == 1)
      return 0;
    int w = (sum + target) / 2;
    vector<int> dp(w + 1, 0);
    dp[0] = 1;
    for (auto num : nums)
      for (int j = w; j >= num; j--)
        dp[j] += dp[j - num];
    return dp[w];
  }
};

[538] 把二叉搜索树转换为累加树

Time: 2022-06-03

Difficulty: Medium (73.65%)

Tags: tree

把链表，树之类的放在数组里就很爽

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
private:
  void dfs(std::vector<TreeNode *> &nodes, TreeNode *root) {
    if (root) {
      nodes.push_back(root);
      dfs(nodes, root->left);
      dfs(nodes, root->right);
    }
  }

public:
  /**
   * @brief
   * 215/215 cases passed (36 ms)
   * Your runtime beats 57.83 % of cpp submissions
   * Your memory usage beats 5.02 % of cpp submissions (33.4 MB)
   * @param root
   * @return TreeNode*
   */
  TreeNode *convertBST(TreeNode *root) {
    std::vector<TreeNode *> nodes;
    dfs(nodes, root);
    std::sort(nodes.begin(), nodes.end(),
              [](TreeNode *a, TreeNode *b) { return a->val < b->val; });
    for (int i = nodes.size() - 2; i >= 0; i--)
      nodes[i]->val += nodes[i + 1]->val;
    return root;
  }
};