【LeetCode018-020】4Sum,单链表Nth去除，符号检测//指针与引用，stack用法

18. 4Sum

 

• Total Accepted: 100564
• Total Submissions: 391517
• Difficulty: Medium
• Contributors: Admin

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: 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]]

class Solution {public:vector<vector<int>> fourSum(vector<int>& nums, int target) {vector<vector<int>> result;vector<int> temp;if(nums.size()<4)return result;sort(nums.begin(), nums.end());//#include<algorithm> for (int i = 0; i<nums.size(); i++) {//四个的剔除掉temp.push_back(nums[i]);}//temp.push_back(nums[nums.size() - 3]);//temp.push_back(nums[nums.size() - 2]);//temp.push_back(nums[nums.size() - 1]);for (int i = 0; i < temp.size(); i++) {for (int j = temp.size() - 1; j > i + 2; j--) {    int begin = i + 1, end = j - 1;while (begin < end) {if(temp[i]+temp[j]+temp[begin]+temp[end]==target){vector<int>M;M.push_back(temp[i]); M.push_back(temp[begin]); M.push_back(temp[end]); M.push_back(temp[j]);if(result.size()==0||result[result.size()-1]!=M)result.push_back(M);    begin++; end--;}if (temp[i] + temp[j] + temp[begin] + temp[end] > target)end--;if (temp[i] + temp[j] + temp[begin] + temp[end] < target)begin++;}                          while(nums[j]==nums[j-1])j--;}    while(nums[i]==nums[i+1])i++;}return result;}};

19. Remove Nth Node From End of List

• Total Accepted: 153567
• Total Submissions: 477873
• Difficulty: Easy
• Contributors: Admin

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.

Note:
Given n will always be valid.
Try to do this in one pass.

class Solution {public:ListNode* removeNthFromEnd(ListNode* head, int n) {ListNode *test;ListNode *result;test = head;result = head;int nums = 0;if(head->next==NULL)return NULL;while (test != NULL) {test = test->next;nums++;}for (int i = 0; i < nums - n-1; i++) {head = head->next;}if (n == nums)result = result->next;if(head->next->next==NULL)head->next = NULL;else head->next = head->next->next;return result;}};

20. Valid Parentheses

• Total Accepted: 161385
• Total Submissions: 502409
• Difficulty: Easy
• Contributors: Admin

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.

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//利用char相减为1或2……，虽然好像很慢…

class Solution {public:    bool isValid(string s) {        bool result=true;        if(s.size()%2==1)return false;        else if(s.size()==0)return true;        else {            int add=1;            for(int i=1;i<s.size();i++){                if(s[i]==s[0])add++;                if(s[i]==s[0]+1||s[i]==s[0]+2){                    add--;                    if(!add){                    result=isValid(s.substr(1,i-1))*isValid(s.substr(i+1,s.size()-i-1));                    break;                    }                }                else{                    result=false;                }            }        }        return result;            }};

结果喜人，击败了0.8%的人，哈哈哈哈,能这么慢也是蛮好玩的

附上优雅的2ms的方法：//stack的用法

class Solution {public:    bool isValid(string s) {        stack<char> paren;        for (char& c : s) {            switch (c) {                case '(':                 case '{':                 case '[': paren.push(c); break;                case ')': if (paren.empty() || paren.top()!='(') return false; else paren.pop(); break;                case '}': if (paren.empty() || paren.top()!='{') return false; else paren.pop(); break;                case ']': if (paren.empty() || paren.top()!='[') return false; else paren.pop(); break;                default: ; // pass            }        }        return paren.empty() ;    }};

去吃饭吃饭~