leetcode 题解代码整理 16-20题

3Sum Closest

Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

    For example, given array S = {-1 2 1 -4}, and target = 1.

    The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

class Solution
{
public:
    int threeSumClosest(vector<int>& nums, int target)
    {
        int ans,i,j,l,r,mid;
        sort(nums.begin(),nums.end());
        ans=20000000;
        for (i=0;i<nums.size()-2;i++)
            for (j=i+1;j<nums.size()-1;j++)
            {
                l=j+1;
                r=nums.size()-1;
            
                if (nums[i]+nums[j]+nums[r]<target)
                {
                    if (abs(nums[i]+nums[j]+nums[r]-target)<abs(ans-target))
                        ans=nums[i]+nums[j]+nums[r];
                    continue;
                }
                if (nums[i]+nums[j]+nums[l]>target)
                {
                    if (abs(nums[i]+nums[j]+nums[r]-target)<abs(ans-target))
                        ans=nums[i]+nums[j]+nums[l];
                    continue;
                }
            
                while (l<=r)
                {
                    mid=(l+r)/2;
                    if (abs(ans-target)>abs(nums[i]+nums[j]+nums[mid]-target))
                        ans=nums[i]+nums[j]+nums[mid];

                    if (ans==target) break;

                    if (nums[i]+nums[j]+nums[mid]<target)
                        l++;
                    else
                        r--;

                }
                if (ans==target) break;
            }
        return ans;
    }
};

 

Letter Combinations of a Phone Number

 

Given a digit string, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below.

Input:Digit string "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

class Solution 
{
vector<string>ans;
string mark;
int len;
public:
    vector<string> letterCombinations(string digits)
    {
        
        len=digits.length();
        ans.clear();
        if (len==0) return ans;
        mark=digits;
        dfs(0,"");
        return ans;
    }
    
private:
    void dfs(int k,string ch)
    {
        if (k==len)
        {
            ans.push_back(ch);
            return ;
        }
        if (mark[k]=='2'){dfs(k+1,ch+"a"); dfs(k+1,ch+"b"); dfs(k+1,ch+"c");}
        if (mark[k]=='3'){dfs(k+1,ch+"d"); dfs(k+1,ch+"e"); dfs(k+1,ch+"f");}
        if (mark[k]=='4'){dfs(k+1,ch+"g"); dfs(k+1,ch+"h"); dfs(k+1,ch+"i");}
        if (mark[k]=='5'){dfs(k+1,ch+"j"); dfs(k+1,ch+"k"); dfs(k+1,ch+"l");}
        if (mark[k]=='6'){dfs(k+1,ch+"m"); dfs(k+1,ch+"n"); dfs(k+1,ch+"o");}
        if (mark[k]=='7'){dfs(k+1,ch+"p"); dfs(k+1,ch+"q"); dfs(k+1,ch+"r"); dfs(k+1,ch+"s");}
        if (mark[k]=='8'){dfs(k+1,ch+"t"); dfs(k+1,ch+"u"); dfs(k+1,ch+"v");}
        if (mark[k]=='9'){dfs(k+1,ch+"w"); dfs(k+1,ch+"x"); dfs(k+1,ch+"y"); dfs(k+1,ch+"z");}
    }
};

4Sum

 

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note:

• Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
• The solution set must not contain duplicate quadruplets.

    For example, given array S = {1 0 -1 0 -2 2}, and target = 0.

    A solution set is:
    (-1,  0, 0, 1)
    (-2, -1, 1, 2)
    (-2,  0, 0, 2)

枚举前两个数字，然后用2sum的思路

class Solution 
{
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) 
    {
        sort(nums.begin(),nums.end());
        
        vector<vector<int> >ans;
        if (nums.size()<4) return ans;
        set<string>st;
        
        
        for (int i=0;i<nums.size()-3;i++)
            for (int j=i+1;j<nums.size()-2;j++)
            {
                int l=j+1;
                int r=nums.size()-1;
                int sum=nums[i]+nums[j];
                while (l<r)
                {
                    int temp=nums[l]+nums[r];
                    if (sum+temp==target)
                    {
                        string str;
                        str+=nums[i];
                        str+=nums[j];
                        str+=nums[l];
                        str+=nums[r];
                        set<string>::iterator it=st.find(str);
                        if (it==st.end())
                        {
                            st.insert(str);
                            vector<int>mark;
                            mark.clear();
                            mark.push_back(nums[i]);
                            mark.push_back(nums[j]);
                            mark.push_back(nums[l]);
                            mark.push_back(nums[r]);
                            ans.push_back(mark);
                        }
                        l++;
                        r--;
                    }
                    else 
                    if (sum+temp<target)
                        l++;
                    else 
                        r--;
                }
            }
        return ans;
        
    }
};

Remove Nth Node From End of List

 

Given a linked list, remove the nth node from the end of list and return its head.

For example,

   Given linked list: 1->2->3->4->5, and n = 2.

   After removing the second node from the end, the linked list becomes 1->2->3->5.

链表处理经典面试题

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (head == NULL)
            return NULL;
            
        ListNode *pPre = NULL;
        ListNode *p = head;
        ListNode *q = head;
        for(int i = 0; i < n - 1; i++)
            q = q->next;
            
        if (q->next==NULL)
        {
            head=head->next;
            return head;
        }
        while(q->next)
        {
            pPre = p;
            p = p->next;
            q = q->next;
        }
        
        
        
            pPre->next = p->next;
            delete p;
        
        
        return head;
    }
};

Valid Parentheses

 

Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.

The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]" and "([)]" are not.

class Solution 
{
public:
    bool isValid(string s) 
    {
        stack<char>st;
        int len=s.length();
        for (int i=0;i<len;i++)
        if (s[i]=='(' || s[i]=='{' || s[i]=='[')
            st.push(s[i]);
        else 
        {
            if (st.size()==0) return false;
            char ch=st.top();
            st.pop();
            if ( (ch=='(' && s[i]==')') || (ch=='[' && s[i]==']') || (ch=='{' && s[i]=='}') );
            else return false;
        }
        
        if (st.size()==0) return true;
        else return false;
    }
};