LeetCode 第 209 场周赛

A 特殊数组的特征值

暴力 模拟

class Solution {
public:
    int specialArray(vector<int>& nums) {
        int n = nums.size();
        for (int i = 1; i <= 100; i++) {
            int m = 0;
            for (int j = 0; j < n; j++) {
                if (nums[j] >= i) m++;
            }
            if (m == i) return i;
        }
        return -1;
    }
};

B 奇偶树

队列 模拟 层序遍历

/**
 * 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 {
public:
    bool isEvenOddTree(TreeNode* root) {
        TreeNode* p;
        queue<TreeNode*> q;
        q.push(root);
        int d = 0;
        while(!q.empty()) {
            int n = q.size();
            int pre = (d % 2 == 0) ? -1 : 1e6 + 10; 
            for (int i = 0; i < n; i++) {
                p = q.front(); q.pop();
                if ((p->val % 2) == (d % 2)) return false;
                if ((d % 2) && pre <= p->val) return false;
                else if(!(d % 2) && pre >= p->val) return false; 
                if (p->left != nullptr) q.push(p->left);
                if (p->right != nullptr) q.push(p->right);
                pre = p->val;
            }
            d++;
        }
        return true;
    }
};

C 可见点的最大数目

class Solution {
public:
    int visiblePoints(vector<vector<int>>& points, int ang, vector<int>& location) {
        int ans = 0;

        vector<double> angle;

        for(const auto &point : points) {
            int x = location[0] - point[0];
            int y = location[1] - point[1];
            if (x == 0 && y == 0) {
                ans++;
            } else {
                angle.emplace_back(atan2(y, x) * 180 / M_PI);
            }
        }

        sort(angle.begin(), angle.end());

        for(int i = 0, n = angle.size(); i < n; i++) {
            angle.emplace_back(angle[i] + 360);
        }

        int anw = ans;
        for(int i = 0, j = 0; i < angle.size(); i++) {
            while(j < angle.size() &&
                (angle[j] - angle[i] < ang + 1e-9)) {
                j++;
            }
            anw = max(j - i + ans, anw);
        }

        return anw;
    }
};

D 使整数变为 0 的最少操作次数

格雷码解码过程(格雷码->二进制码): 从左边第二位起, 将每位与左边一位解码后的值异或, 作为该位解码后的值

class Solution {
public:
    int minimumOneBitOperations(int n) {
        int ans = 0;
        while (n) {
            ans ^= n;
            n >>= 2;   
        } 
        return ans;
    }
};