leetcode 题解代码整理 16-20题

3Sum Closest

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;
}
};