每日算法刷题Day13-在O(1)时间删除链表结点、合并两个排序的链表、把字符串转换成整数

输入：链表 1->4->6->8
      删掉节点：第2个节点即6（头节点为第0个节点）

输出：新链表 1->4->8

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    void deleteNode(ListNode* node) {
        node -> val = node -> next ->val;
        node -> next = node -> next ->next;
    }
};

class Solution {
public:
    void deleteNode(ListNode* node) {
        *(node) = *(node -> next);
    }
};

输入：1->3->5 , 2->4->5

输出：1->2->3->4->5->5

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* merge(ListNode* l1, ListNode* l2) {
        auto dummy = new ListNode(-1),tail = dummy;
        while(l1 && l2)
            if(l1 -> val < l2 -> val)
            {
                tail = tail -> next = l1;
                l1 = l1 -> next;
            }
            else
            {
                tail = tail -> next = l2;
                l2 = l2 -> next;
            }
        if(l1)tail -> next = l1;
        if(l2)tail -> next = l2;
        
        return dummy ->next;
    }
};

atoi(str.c_str())

输入："123"

输出：123

class Solution {
public:
    int strToInt(string str) {
        int k = 0;
        long long f = 1;
        //处理正负
        while (k < str.size() && str[k] == ' ')k++;
        
        if(k < str.size() && str[k] == '-')
        {
            k++;
            f = -1;
        }
        if( k < str.size() && str[k] == '+')k++;
        
        
        long long r = 0;
        while(k < str.size() && str[k] <= '9' && str[k] >= '0')
        {
            //如果是负数，则每一位转换为负数最后相加即可
            //通过乘以10移位
            r = r*10 + f * (str[k] - '0');
            if (r < INT_MIN) return INT_MIN;
            if (r > INT_MAX) return INT_MAX;
            k ++;
        }
        return (int)r;
    }
};