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;    }};