LeetCode每日1题——2023.03

周日	周一	周二	周三	周四	周五	周六
			1 √	2 √	3 √	4 位运算 ×
5 √	6 思维 ×	7 栈 ×	8 √	9 √	10 √	11 √
12 √	13 √	14 贪心 ×	15 √	16 前缀和 枚举 ×	17 √	18 双指针 ×
19 √	20 数位DP ×	21 √	22 DP ×	23 √	24 字典树 ×	25 √
26 √	27 √	28 DP √	29 √	30 √	31 √

统计每个月的每日1题做题情况。统计规则为：

• 若正常解题且不为暴力做法，则日期后做标记 √
• 若正常解题且为暴力做法，则日期后做标记 √
• 若无法解题，则日期后做标记 ×

1 矩阵中的局部最大值

链接

代码

class Solution {
public:
    vector<vector<int>> largestLocal(vector<vector<int>>& grid) {
        vector<vector<int>> ans;
        int n  = grid.size();
        int dy[8] = {-1, 0, 1, -1, 1, -1, 0, 1};
        int dx[8] = {-1, -1, -1, 0, 0, 1, 1, 1};
        for(int i = 1; i < n-1; i++){
            vector<int> tmp;
            for(int j = 1; j < n-1; j++){
                int now_max = grid[i][j];
                for(int k = 0; k < 8; k++)  now_max = max(grid[i+dx[k]][j+dy[k]], now_max);
                tmp.push_back(now_max);
            }
            ans.push_back(tmp);
        }
        return ans;
    }
};

2 二进制数转字符串

链接

代码

class Solution {
public:
    string printBin(double num) {
        string s ="0.";
        double eps = 1e-6;
        int cnt = 2;
        while(num > eps && cnt++ <= 32){
            num *= 2;
            if(num >= 1) {
                num -= 1;
                s += "1";
            }
            else s += "0";
            cnt++;
        }
        if(cnt >= 32)
            return "ERROR";
        else 
            return s;
    }
};

3 保证文件名唯一

链接

代码

class Solution {
public:
    string toString(string s, int num){
        stringstream ss;
        ss << num;
        string tmp;
        ss >> tmp;
        return s +"(" + tmp + ")";
    }
    vector<string> getFolderNames(vector<string>& names) {
        set<string> st;
        map<string, int> mp;
        vector<string> ans;
        for(int i = 0; i < names.size(); i++){
            if(st.find(names[i]) == st.end()){
                ans.push_back(names[i]);
                st.insert(names[i]);
            }
            else {
                int cnt = mp[names[i]];
                while(++cnt){
                    string tmp = toString(names[i], cnt);
                    if(st.find(tmp) == st.end()){
                        ans.push_back(tmp);
                        st.insert(tmp);
                        mp[names[i]] = cnt;
                        break;
                    }
                }
            }
        }
        return ans;
    }
};

4 按位与为零的三元组

链接

代码

class Solution {
public:
    int countTriplets(vector<int>& nums) {
        int n = nums.size();
        int x[1 << 16] = {0};
        int sum = 0;
        for(int i = 0; i < n; i++){
            for(int j = 0; j < n; j++)
                ++x[nums[i] & nums[j]];
        }
        for(int i = 0; i < n; i++){
            for(int j = 0; j < (1<<16); j++){
                if((j & nums[i]) == 0) sum += x[j];
            }
        }
        return sum;
    }
};

5 经营摩天轮的最大利润

链接

代码

class Solution {
public:
    int minOperationsMaxProfit(vector<int>& customers, int boardingCost, int runningCost) {
        int max_value = -1000;
        int sum = 0;
        int index = 0;
        int n = customers.size();
        for(int i = 0; i < n; i++){
            if(customers[i] >= 4){
                sum += 4*boardingCost - runningCost;
                if(i != n-1) customers[i+1] += customers[i] - 4;
            }
            else sum += customers[i]*boardingCost - runningCost;
            if(sum > max_value){
                max_value = sum;
                index = i;
            }
        }
        int tmp = customers[n-1];
        int tmp_index = n;
        while(tmp){
            if(tmp >= 4){
                sum += 4*boardingCost - runningCost;
                tmp -= 4;
            }
            else {
                sum += tmp*boardingCost - runningCost;
                tmp = 0;
            }
            if(sum > max_value){
                max_value = sum;
                index = tmp_index;
            }
            tmp_index++;
        }
        if(max_value <= 0) 
            return -1;
        else
            return index;
    }
};

6 使字符串平衡的最少删除次数

链接

思路

代码

class Solution {
public:
    int minimumDeletions(string s) {
        int n = s.length();
        int left_b = 0;
        int right_a = 0;
        for(int i = 0; i < n; i++){
            if(s[i] == 'a') ++right_a;
        }
        int res = right_a;
        for(int i = 0; i < n; i++){
            if(s[i] == 'a') --right_a;
            else ++left_b;
            res = min(res, left_b+right_a);
        }
        return res;
    }
};

7 花括号展开 II

链接

思路

代码

class Solution {
public:
    void operate(vector<char> &op, vector<set<string>> &stk){
        int left = stk.size() - 2;
        int right = stk.size() - 1;
        if(op.back() == '+') {
            stk[left].merge(stk[right]);
        }
        else {
            set<string> tmp;
            for(auto &l: stk[left])
                for(auto &r: stk[right]){
                    tmp.insert(l+r);
                }
            stk[left] = move(tmp);
        }
        op.pop_back();
        stk.pop_back();
    }
    vector<string> braceExpansionII(string expression) {
        vector<char> op;
        vector<set<string>> stk;

        for(int i = 0; i < expression.size(); i++){
            if(expression[i] == ','){
                while(op.size() && op.back() == '*') operate(op, stk);
                op.push_back('+');
            }
            else if(expression[i] == '{'){
                if(i > 0 && (isalpha(expression[i-1]) || expression[i-1] == '}')) op.push_back('*');
                op.push_back(expression[i]);
            }
            else if(expression[i] == '}'){
                while(op.size()) {
                    operate(op, stk);
                    if(op.back() == '{'){
                        op.pop_back();
                        break;
                    }
                }
            }
            else {
                if(i > 0 && (isalpha(expression[i-1]) || expression[i-1] == '}')) op.push_back('*');
                stk.push_back({string(1, expression[i])});
            }
        }
        
        while(op.size()) {
            operate(op, stk);
        }
        return {stk.back().begin(), stk.back().end()};
    }
};

8 礼物的最大价值

链接

代码

class Solution {
public:
    int maxValue(vector<vector<int>>& grid) {
        const int m = grid.size();
        const int n = grid[0].size();
        
        int dp[m][n];
        dp[0][0] = grid[0][0];
        for(int i = 1; i < n; i++) dp[0][i] = grid[0][i] + dp[0][i-1];
        for(int j = 1; j < m; j++) dp[j][0] = grid[j][0] + dp[j-1][0];
        for(int i = 1; i < m; i++) 
            for(int j = 1; j < n; j++) 
                dp[i][j] = grid[i][j] + max(dp[i-1][j], dp[i][j-1]);
        return dp[m-1][n-1];
    }
};

9 得到 K 个黑块的最少涂色次数

链接

代码

class Solution {
public:
    int minimumRecolors(string blocks, int k) {
        int left = 0;
        int right = 0;
        int cnt = 0;
        while(right < k) {
            cnt += (blocks[right++]=='W')?1:0;           
        }
        int ans = cnt;
        while(right < blocks.size()) {
            cnt += (blocks[right++]=='W')?1:0;
            cnt -= (blocks[left++]=='W')?1:0;
            ans = min(ans, cnt);       
        }

        return ans;
    }
};

10 使数组和能被 P 整除

链接

代码

class Solution {
public:
    int minSubarray(vector<int>& nums, int p) {
        int res = 0;
        for(int i = 0; i < nums.size(); i++) 
            res = (res + nums[i]) % p;
        if(res == 0) return 0;
        map<int, int> mp; //  = <余数， 下标>
        mp[0] = -1; //下标为-1代表前缀和为空，空即为0
        int sum = 0;
        int ans = nums.size();
        for(int i = 0; i < nums.size(); i++){
            sum = (sum + nums[i]) % p;
            mp[sum] = i;
            if(mp.count((sum - res + p) % p) > 0){
                ans = min(ans, i - mp[(sum - res + p) % p]);
            }
        }
        return ans == nums.size()? -1: ans;   
    }
};

11 字母与数字

链接

代码

class Solution {
public:
    vector<string> findLongestSubarray(vector<string>& array) {
        int n = array.size();
        int presum[n+1];
        int max_line = 0;
        int max_left = 100000;
        int max_right = -1;
        map<int, int> mp;
        set<int> st;
        st.insert(0);
        mp[0] = 0;
        presum[0] = 0;
        for(int i = 0; i < n; i++){
            if(isalpha(array[i][0])) presum[i+1] = presum[i] - 1;
            else presum[i+1] = presum[i] + 1;
            if(!st.count(presum[i+1])) {
                mp[presum[i+1]] = i;
                st.insert(presum[i+1]);
            }
            int left = mp[presum[i+1]];
            if(!presum[i+1] && i+1 > max_line) {
                max_left = 0;
                max_right = i;
                max_line = i+1;
            }
            else if(i - left > max_line) {
                max_left = left+1;
                max_right = i;
                max_line = i - left;
            }
        }
        vector<string> ans;
        if(max_line) {
            auto left = array.begin() + max_left;
            auto right = array.begin() + max_right + 1;
            ans.assign(left, right);
        } 
        return ans;
    }
};

12 统计子树中城市之间最大距离

链接

代码

class Solution {
private:
    int g[20][20];
public:
    void dfs(int vis[], int point, int n){
        vis[point] = 0;
        for(int i = 1; i <= n; i++){
            if(vis[i] && g[point][i] != n+1)
                dfs(vis, i, n);
        }
    }
    vector<int> countSubgraphsForEachDiameter(int n, vector<vector<int>>& edges) {
        int vis[n+5];
        int mark[n+5];
        vector<int> ans(n-1);
        for(int i = 0; i < n-1; i++) ans[i] = 0;
        
        for(int i = 1; i < (1 << n); i++){

            int start;
            for(int j = 0; j < n ; j++) {
                vis[j+1] = (i >> j) % 2;
                mark[j+1] = vis[j+1];
                if(vis[j+1]) start = j+1;
            }
            
            for(int k = 1; k <= n; k++)
                for(int j = 1; j <= n; j++) {
                    if(k != j) g[k][j] = n+1;
                    else g[k][j] = 0;
                }

            for(int k = 0; k < edges.size(); k++) {
                if(vis[edges[k][0]] && vis[edges[k][1]])
                    g[edges[k][0]][edges[k][1]] = g[edges[k][1]][edges[k][0]] = 1;
            }
            
            dfs(mark, start, n);
            int v = 0;
            for(int i = 1; i <= n; i++){
                if(mark[i]){
                    v = 1;
                    break;
                }
            }
            if(v) continue;

            for(int x = 1; x <= n; x++){
                for(int y = 1; y <= n; y++)
                    for(int z = 1; z <= n; z++)
                        g[y][z] = min(g[y][z], g[y][x] + g[x][z]);
            }


            int max_edge = -1;
            int flag = 1;
            for(int y = 1; y <= n; y++)
                for(int z = 1; z <= n; z++){
                    if(vis[y] && vis[z] && y != z) {
                        if(g[y][z] == n+1){
                            flag = 0;
                            break;
                        }
                        else max_edge = max(max_edge, g[y][z]);
                    }
                }
            
            if(flag && max_edge != -1)
                ans[max_edge-1]++;
            
        }
        return ans;
    }
};

13 赢得比赛需要的最少训练时长

链接

代码

class Solution {
public:
    int minNumberOfHours(int initialEnergy, int initialExperience, vector<int>& energy, vector<int>& experience) {
        int n = energy.size();
        int sum_energy = 1;
        int sum_experience = 0;
        for(int i = 0; i < n; i++){
            sum_energy += energy[i];
            if(experience[i] >= initialExperience){
                sum_experience += experience[i] - initialExperience + 1;
                initialExperience += experience[i] - initialExperience + 1;;
            }
            initialExperience += experience[i];
        }
        int tmp;
        if(sum_energy - initialEnergy > 0) tmp = sum_energy - initialEnergy;
        else tmp = 0;
        return (tmp + sum_experience > 0)?tmp + sum_experience:0;
    }
};

14 给定行和列的和求可行矩阵

链接

代码

class Solution {
public:
    vector<vector<int>> restoreMatrix(vector<int>& rowSum, vector<int>& colSum) {
        vector<vector<int>> ans;
        int n = rowSum.size();
        int m = colSum.size();
        for(int i = 0; i < n; i++){
            vector<int> tmp;
            for(int j = 0; j < m; j++){
                int val = min(rowSum[i], colSum[j]);
                tmp.push_back(val);
                rowSum[i] -= val;
                colSum[j] -= val;
            }
            ans.push_back(tmp);
        }
        return ans;
    }
};

15 最大网络秩

链接

代码

class Solution {
public:
    int maximalNetworkRank(int n, vector<vector<int>>& roads) {
        int g[n][n];
        memset(g, 0, sizeof(g));
        for(int i = 0; i < roads.size(); i++)
            g[roads[i][0]][roads[i][1]] = g[roads[i][1]][roads[i][0]] = 1;
        int ans = 0;
        for(int i = 0; i < n; i++)
            for(int j = 0; j < n; j++) {
                if(i != j){
                    int tmp_sum = 0;
                    for(int k = 0; k < n; k++) {
                        tmp_sum += g[i][k] + g[j][k];
                    }
                    // cout << " i = " << i << " j = " << j << " tmp_sum = " << tmp_sum << endl;
                    if(g[i][j]) tmp_sum -= 1;
                    ans = max(ans, tmp_sum);
                }
            }
        return ans;

    }
};

16 统计中位数为 K 的子数组

链接

思路

代码

class Solution {
public:
    inline int trans(int num){
        if(num == 0) return 0;
        else return (num > 0)? 1:-1;
    }
    int countSubarrays(vector<int>& nums, int k) {        
        int ans_index = 0;
        int n = nums.size();
        for(int i = 0; i < n; i++){
            if(nums[i] == k) {
                ans_index = i;
                break;
            }
        }
        int pre_sum = 0;
        int ans = 0;
        map<int, int> mp;
        mp[0] = 1;
        for(int i = 0; i < n; i++){
            pre_sum += trans(nums[i] - k);
            if(i < ans_index)
                mp[pre_sum]++;
            else {
                int pre0 = mp[pre_sum];
                int pre1 = mp[pre_sum-1];
                ans += pre0 + pre1;
            }
        }
        return ans;
    }
};

17 和有限的最长子序列

链接

代码

class Solution {
public:
    vector<int> answerQueries(vector<int>& nums, vector<int>& queries) {
        int n = nums.size();
        int m = queries.size();
        vector<int> pre_sum(n+1);
        pre_sum[0] = 0;
        sort(nums.begin(), nums.end());
        for(int i = 0; i < n; i++)
            pre_sum[i+1] = pre_sum[i] + nums[i];
        vector<int> ans(m);
        for(int i = 0; i < m; i++){
            ans[i] = upper_bound(pre_sum.begin(), pre_sum.end(), queries[i]) - pre_sum.begin() -1;
        }
        return ans;
    }
};

18 分割两个字符串得到回文串

链接

代码

class Solution {
public:
    bool checkSplitPoint(string &a, int left, int right) {
        while(left < right && a[left] == a[right]){
            left++;
            right--;
        } 
        return left >= right;
    }
    bool TwoPointCheck(string &a, string &b) {
        int n = a.length();
        int left = 0;
        int right = n-1;
        while(left < right && a[left] == b[right]){
            left++;
            right--;
        }
        if(left >= right) return true;
        return checkSplitPoint(a, left, right) ||  checkSplitPoint(b, left, right);
    }
    bool checkPalindromeFormation(string a, string b) {
        return TwoPointCheck(a, b) || TwoPointCheck(b, a);
    }
};

19 执行操作后字典序最小的字符串

链接

代码

class Solution {
public:
    string findLexSmallestString(string s, int a, int b) {
        string ans = s;
        int n = s.length();
        queue<string> q;
        set<string> st;
        st.insert(s);
        q.push(s);
        int cnt = 0;
        while(!q.empty()){
            string tmp_aop = q.front();
            string tmp_bop = tmp_aop;
            q.pop();
            for(int l = 1; l < n; l+=2){
                tmp_aop[l] = ((tmp_aop[l] - '0') + a) % 10 + '0';
            }
            for(int k = 0; k < n; k++){
                tmp_bop[k] = tmp_aop[(n-b+k)%n];
            }
            if(!st.count(tmp_aop)){
                q.push(tmp_aop);
                st.insert(tmp_aop);
            }
            if(!st.count(tmp_bop)){
                q.push(tmp_bop);
                st.insert(tmp_bop);
            }
            ans = min(ans, tmp_bop);
            ans = min(ans, tmp_aop);
        }
        return ans;
    }
};

20 至少有 1 位重复的数字

链接

思路

代码

class Solution {
private:
    int dp[11][1<<11];
public:
    int dfs(int i, int mask, string s, bool same){
        if(i == s.length()) // 取数完成，返回1 
            return 1;
        if(!same && dp[i][mask]) // 当前所取数小于n, 并且之前已经生成相关子树，则直接返回
            return dp[i][mask];
        int up = same?(s[i] - '0'):9;
        int res = 0;
        for(int k = 0; k <= up; k++){
            if(mask & (1 << k)) continue;
            res += dfs(i+1, (!mask && !k)?0:mask|(1<<k), s, same && k == up);
        }
        if(!same)
            dp[i][mask] = res;
        return res;
    }
    int numDupDigitsAtMostN(int n) {
        for(int i = 0; i < 11; i++)
            for(int j = 0; j < (1<<11); j++) dp[i][j] = 0;
        string s = to_string(n);
        return n + 1 - dfs(0, 0, s, true);
    }
};

21 温度转换

链接

代码

class Solution {
public:
    vector<double> convertTemperature(double celsius) {
        vector<double> ans;
        ans.push_back(celsius+273.15);
        ans.push_back(celsius*1.80 + 32.00);
        return ans;
    }
};

22 无矛盾的最佳球队

链接

代码

class Solution {
public:
    struct player{
        int scores;
        int ages;
    };
private:
    player p[1005];
public:
    static bool cmp(player p1, player p2){
        if(p1.ages != p2.ages)
            return p1.ages < p2.ages;
        else 
            return p1.scores < p2.scores;
    }
    int bestTeamScore(vector<int>& scores, vector<int>& ages) {
        int n = scores.size();
        int ans = 0;
        for(int i = 0; i < n; i++){
            p[i].scores = scores[i];
            p[i].ages = ages[i];
        }
        sort(p, p+n, cmp);
        int dp[n];
        for(int i = 0; i < n; i++){
            dp[i] = p[i].scores;
            for(int j = 0; j < i; j++){
                if(p[j].scores <= p[i].scores)
                    dp[i] = max(dp[i], dp[j] + p[i].scores);
            }
        }
        for(int i = 0; i < n; i++) ans = max(ans, dp[i]);
        return ans; 
    }
};

23 等差子数组

链接

代码

class Solution {
public:
    vector<bool> checkArithmeticSubarrays(vector<int>& nums, vector<int>& l, vector<int>& r) {
        vector<bool> ans;
        for(int i = 0; i < l.size(); i++){
            vector<int> tmp(nums.begin() + l[i], nums.begin() + r[i]+1);
            sort(tmp.begin(), tmp.end());
            int flag = 1;
            int gap = tmp[1] - tmp[0];
            for(int i = 2; i < tmp.size(); i++) {
                if(tmp[i] - tmp[i-1] != gap) {
                    flag = 0; break;
                }
            }
            if(flag) ans.push_back(true);
            else ans.push_back(false);
        }
        return ans;
    }
};

24 字符流

链接

代码

class Tire{
private:
    vector<Tire *> children;
    bool is_End;
    
public:
    Tire():children(26), is_End(false){}

    void insert(string &w){
        Tire *p = this;
        reverse(w.begin(), w.end());
        for(char &c: w){
            int idx = c - 'a';
            if(!p->children[idx]){
                p->children[idx] = new Tire();
            }
            p = p->children[idx];
        }
        p->is_End = true;
    }

    bool search(string &s){
        Tire *p = this;
        for(int i = s.length()-1, j = 0; i >= 0 && j < 201; i--, j++){
            int idx = s[i] - 'a';
            if(p->is_End){
                return true;
            }
            if(!p->children[idx]){
                return false;
            }
            p = p->children[idx];
        }
        return p->is_End;
    }
};

class StreamChecker {
public:
    Tire *t = new Tire();
    string s;
public:
    StreamChecker(vector<string>& words) {
        for(string &w: words)
            t->insert(w);
    }
    
    bool query(char letter) {
        s += letter;
        return t->search(s);
    }
};

25 删除最短的子数组使剩余数组有序

链接

代码

class Solution {
public:
    int findLengthOfShortestSubarray(vector<int>& arr) {
        int n = arr.size();
        int i = 0, j = n - 1;
        while(j > 0 && arr[j] >= arr[j-1]) j--;
        if(j == 0) return 0;
        int res = j;
        for(; i < n; i++){
            while(j < n && arr[j] < arr[i]) j++;
            res = min(res, j-i-1);
            if(i+1 < n && arr[i] > arr[i+1]) break;
        }
        return res;
    }
};

26 和相等的子数组

链接

代码

class Solution {
public:
    bool findSubarrays(vector<int>& nums) {
        int windows = 2;
        int n = nums.size();
        map<int, int> mp;
        for(int i = 0; i+1 < n; i++){
            if(mp.count(nums[i]+nums[i+1]) > 0 && mp[nums[i]+nums[i+1]] != i)
                return true;
            if(!mp.count(nums[i]+nums[i+1]))
                mp[nums[i]+nums[i+1]] = i;
        }
        return false;
    }
};

27 统计只差一个字符的子串数目

链接

代码

class Solution {
public:
    int countSubstrings(string s, string t) {
        int windows = 0;
        int n = s.length();
        int m = t.length();
        int ans = 0;
        while(windows++ < n){
            for(int i = 0; i + windows <= n; i++){
                for(int j = 0; j + windows <= m; j++){
                    int diff = 0;
                    for(int k = 0; k < windows; k++){
                       if(diff > 1) break;
                       if(s[i+k] != t[j+k]) diff++;
                    }
                    if(diff == 1) ans++;
                }
            }
        }
        return ans;
    }
};

28 最短公共超序列

链接

代码

class Solution {
public:
    string shortestCommonSupersequence(string str1, string str2) {
        int n = str1.length();
        int m = str2.length();
        int dp[m+1][n+1];

        for(int i = 0; i <= n; i++) dp[0][i] = i;
        for(int i = 0; i <= m; i++) dp[i][0] = i;
       
        for(int i = 1; i <= m; i++){
            for(int j = 1; j <= n; j++){
                if(str1[j-1] == str2[i-1]) 
                    dp[i][j] = dp[i-1][j-1] + 1;
                else
                    dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + 1;

            }
        }
        
        int i = m, j = n;
        string ans;
        while(i > 0 && j > 0){
            if(str1[j-1] == str2[i-1]){
                ans += str2[i-1]; j--; i--;    
            }
            else {
                if(i > 0 && dp[i][j] == dp[i-1][j] + 1){
                    ans += str2[i-1]; i--;
                }
                else{
                    ans += str1[j-1]; j--;
                }
            }
        }
        while(i > 0){ans += str2[i-1];i--;}
        while(j > 0){ans += str1[j-1];j--;}
        reverse(ans.begin(), ans.end());
        return ans;
    }
};

29 统计字典序元音字符串的数目

链接

代码

class Solution {
public:
    int countVowelStrings(int n) {
        int dp[n][5];
        for(int i = 0; i < n; i++)
            for(int j = 0; j < 5; j++)
                dp[i][j] = 0;
        for(int i = 0; i < 5; i++) dp[0][i] = 1;
        for(int i = 1; i < n; i++)
            for(int j = 0; j < 5; j++)
                for(int k = 4; k >= j; k--)
                    dp[i][j] += dp[i-1][k];
        int sum = 0;
        for(int i = 0; i < 5; i++) sum += dp[n-1][i];
        return sum;
    }
};

30 两点之间不包含任何点的最宽垂直区域

链接

代码

class Solution {
public:
    struct point{
        int x;
        int y;
    };
    static bool cmp(point p1, point p2){
        return p1.x < p2.x;
    }
    int maxWidthOfVerticalArea(vector<vector<int>>& points) {
        point p[100005];
        int n = points.size();
        for(int i = 0; i < n; i++){
            p[i].x = points[i][0];
            p[i].y = points[i][1];
        }
        sort(p, p+n, cmp);
        int ans = p[1].x - p[0].x;
        for(int i = 1; i < n-1; i++)
            ans = max(ans, p[i+1].x - p[i].x);
        return ans;
    }
};

31 算术三元组的数目

链接

代码

class Solution {
public:
    int arithmeticTriplets(vector<int>& nums, int diff) {
        int sum = 0;
        int n = nums.size();
        for(int i = 0; i < n; i++){
            for(int j = i+1; j < n; j++)
                for(int k = j+1; k < n; k++)
                    if(nums[j] - nums[i] == diff && nums[k] - nums[j] == diff)
                        sum++;
        }
        return sum;
    }
};