1.有序数组问题

1.1 leetcode 15. 三数之和

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class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
    int len = nums.size();
    vector< vector<int> > res;
    sort(nums.begin(),nums.end());
    for(int i = 0;i < len;i++)
    {
    	int opp = -nums[i];
    	if(i==0 || nums[i] != nums[i-1])
    	{
    		int left = i+1,right = len-1;
    		while(left < right)
    		{
    			int sum = nums[left] + nums[right];
    			if(sum == opp)
    			{
    				res.push_back({nums[i],nums[left],nums[right]});
    				while(left < right && nums[left] == nums[left+1])++left;++left;
                    while(left < right && nums[right] == nums[right-1])--right;--right;
    			}
    			else if(sum < opp)
    			{
    				left++;
    			}
    			else 	right--;
    		}
    	}
    }
    return res;
    }
};

1.2 16. 最接近的三数之和

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class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        int _min = 0x3f3f3f3f;
        int ans;
        sort(nums.begin(),nums.end());
        int len = nums.size();
        for(int i = 0;i<len;i++)
        {
        	if(i > 0 && nums[i] == nums[i-1])continue;
        	int left = i+1,right = len-1;
        	while(left < right)
        	{
        		if(nums[i] + nums[left] + nums[right] == target)
        		{
        			return target;
        		}
        		else if (nums[i] + nums[left] + nums[right] < target)
        		{
        			
        			if(abs(nums[i] + nums[left] + nums[right] - target) < _min)
        				ans = nums[i] + nums[left] + nums[right] ,_min = abs(nums[i] + nums[left] + nums[right] - target);
        			left++;
        		}
        		else {
        			if(abs(nums[i] + nums[left] + nums[right] - target) < _min)
        				ans = nums[i] + nums[left] + nums[right] ,_min = abs(nums[i] + nums[left] + nums[right] - target);
        			right--;
        		}
        	}
        }
        return ans;
    }
};

1.3 18. 四数之和

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class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        int len = nums.size();
        vector < vector<int> >res;
        sort(nums.begin(),nums.end());
        for(int i = 0;i<len;i++)
        {
        	if(i>0 && nums[i] == nums[i-1])continue;
        	for(int j = i+1;j<len;j++)
        	{
        		if(j > i+1 && nums[j] == nums[j-1])continue;
        		int left = j+1,right = len-1;
        		while(left < right)
        		{
        			if(nums[i] + nums[j] + nums[left] + nums[right] == target)
        			{
        				res.push_back({nums[i],nums[j],nums[left],nums[right]});
        				while(left < right && nums[left] == nums[left+1])left++;
        				while(left < right && nums[right] == nums[right-1])right--;
        				left++;right--;

        			}
        			else if(nums[i] + nums[j] + nums[left] + nums[right] < target)
        			{
        				left++;
        			}
        			else right--;
        		}
        	}
        }
        return res;
    }
};

1.4 11. 盛最多水的容器

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class Solution {
public:
    int maxArea(vector<int>& height) {
        int l = 0, r = height.size()-1,_max = -1;
        while(l<r)
        {
            _max = max(_max,min(height[l],height[r])*(r-l));
            if(height[l] < height[r])++l;
            else --r;    
        }
        return _max;
    }
};

2.链表问题

141. 环形链表

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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    bool hasCycle(ListNode *head) {
        if(!head || !head->next)return false;
        auto fast = head->next->next,slow = head->next;
        
        while(fast != slow)
        {
            if(!fast || !slow) return false;
            slow = slow->next;
            if(!fast->next)return false;
            fast = fast->next->next;
        }
        return true;
    }
};

这道题还有一个就是需要找到环的入口，可以设一下变量，轻松推出来做法。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        if(!head || !head->next)return NULL;
        
        auto slow = head->next,fast = head->next->next;
        while(slow != fast)
        {
            if(!slow || !fast || !fast->next)
                return NULL;
            
            fast = fast->next->next;
            slow = slow->next;
        }
        fast = head;
        while(slow!=fast)
        {
            slow = slow->next;
            fast = fast->next;
        }
        return fast;
        
    }
};

19. 删除链表的倒数第N个节点

/**
 * 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) {
        if(head==nullptr)return nullptr;
        ListNode *i = head, *ii = head;
        while(n && ii->next){
        	ii = ii->next;
        	n--;
        }
        if(n==1)return head->next;
        
        while(ii->next)
        {
        	i = i->next;
        	ii = ii->next;
        }
        i->next = i->next-> next;
        return head;
    }
};

两个链表的第一个公共结点

/*
struct ListNode {
	int val;
	struct ListNode *next;
	ListNode(int x) :
			val(x), next(NULL) {
	}
};*/
class Solution {
public:
    ListNode* FindFirstCommonNode( ListNode* pHead1, ListNode* pHead2) {
        int n = 0, m = 0;
        auto p = pHead1;
        while(p){
            n++;
            p = p->next;
        }
        p = pHead2;
        while(p){
            m++,p = p->next;
        }
        auto p1 = pHead1,p2 = pHead2;
        while(n!=m){
            if(n>m){
                p1 = p1->next;n--;
            }
            else{
                p2 = p2->next;m--;
            }
            
            
        }
        for(int i = 0;i<n;i++){
            if(p1->val == p2->val)return p1;
            p1 = p1->next,p2 = p2->next;
        }
        
        return nullptr;
    }
};