LeetCode每日一题

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160. 相交链表

原题链接:相交链表

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        vector<ListNode *> va,vb;
        ListNode * headA_tmp = headA,*headB_tmp = headB;
         while(headA_tmp!=nullptr){
            va.push_back(headA_tmp);
            headA_tmp = headA_tmp->next;
        }
         while(headB_tmp!=nullptr){
            vb.push_back(headB_tmp);
            headB_tmp = headB_tmp->next;
        }
        if(va.back()!=vb.back()) return nullptr;
        int i = va.size()-2, j = vb.size()-2;
        while(i>=0&&j>=0){
            if(va[i]!=vb[j])
                break;
            i--;
            j--;
        }
        return va[i+1];
    }
};

402. 移掉 K 位数字

原题链接:移掉 K 位数字

class Solution {
public:
    string minnum = "";
    void dfs(string num,string s,int begin,int len)
    {
        if(s.size()==len)
        {
            int num1 = std::atoi(s.c_str());
            int num2 = std::atoi(minnum.c_str());
            if(num1<num2)
                minnum = s;
            return;
        }
        for(int i = begin;i<num.size();i++)
        {
            dfs(num,s+num[i],i+1,len);
        }
    }
    string removeKdigits(string num, int k) {
        // minnum = num;
        // string s = "";
        // int len = num.size() - k;
        // dfs(num,s,0,len);
        // minnum = to_string(atoi(minnum.c_str()));
        // return minnum;
        vector<char> stk;
        for(auto ch:num){
            while(stk.size()>0&&stk.back()>ch&&k)
            {
                stk.pop_back();
                k--;
            }
            stk.push_back(ch);
        }
        while(k--)
        {
            stk.pop_back();//将剩余没有pop掉的给pop
        }
        string ans = "";
        bool isLeadingZero = true;
        for (auto& digit: stk) {
            if (isLeadingZero && digit == '0') {
                continue;
            }
            isLeadingZero = false;
            ans += digit;
        }
        return ans == "" ? "0" : ans;

    }
};

141. 环形链表

原题链接: 环形链表

class Solution {
public:
    bool hasCycle(ListNode *head) {
        set<ListNode*> st;
        int sz = st.size();
        while(head){
            st.insert(head);
            if(sz==st.size())
            {
                return true;
            }
            sz = st.size();
            head = head->next;
        }
        return false;
    }
};

快慢指针：

class Solution {
public:
    bool hasCycle(ListNode* head) {
        if (head == nullptr || head->next == nullptr) {
            return false;
        }
        ListNode* slow = head;
        ListNode* fast = head->next;
        while (slow != fast) {
            if (fast == nullptr || fast->next == nullptr) {
                return false;
            }
            slow = slow->next;
            fast = fast->next->next;
        }
        return true;
    }
};

主持人调度

原题链接:主持人调度

class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param schedule int整型vector<vector<>> 
     * @return bool布尔型
     */
    bool hostschedule(vector<vector<int> >& schedule) {
        // write code here
        //进行升序排序
        sort(schedule.begin(),schedule.end(),[](const vector<int>& v1,const vector<int>& v2){
            return v1[0]<v2[0];
        });
        int prev = schedule[0][1];
        for(int i = 1;i<schedule.size();i++)
        {
          
                if(schedule[i][0]<prev)
                    return false;
                prev = schedule[i][1];
            
        }
        return true;
    }
};

有效三角形的个数

原题链接: 有效三角形的个数

class Solution {
public:
    // int count = 0;
    // void dfs(vector<int> group,vector<int>& nums,int pos)
    // {
    //     if(group.size()==3)
    //     {
    //         if(group[0]+group[1]>group[2]&&group[0]+group[2]>group[1]&&group[1]+group[2]>group[0])
    //             count++;
    //         return;
    //     }
    //     for(int i = pos;i<nums.size();i++)
    //     {
    //         group.push_back(nums[i]);
    //         dfs(group,nums,i+1);
    //         group.pop_back();
    //     }
    // }
    int triangleNumber(vector<int>& nums) {
        // vector<int> group;
        // dfs(group,nums,0);
        // return count;
        
		//排序+二分查找
        int n = nums.size();
        sort(nums.begin(),nums.end());
        int res = 0;
        for(int i = 0;i<n;i++)
        {
          for(int j = i+1;j<n;j++)
          {
               int left = j+1,right = n-1,k = j;
                while(left<=right)
                {
                    int mid = (left+right)/2;
                    if(nums[mid]<nums[i]+nums[j])
                    {
                        k = mid;
                        left = mid + 1;
                    }
                    else
                    {
                        right = mid-1;
                    }
                }
                res+=k-j;
          }
        }
        return res;
    }
};