[日常刷题]leetcode第四天

20. Valid Parentheses

Given a string containing just the characters ‘(‘, ‘)’, ‘{‘, ‘}’, ‘[’ and ‘]’, determine if the input string is valid.

An input string is valid if:

1.Open brackets must be closed by the same type of brackets.
2.Open brackets must be closed in the correct order.

Note that an empty string is also considered valid.

Example 1:

Input: "()"
Output: true

Example 2:

Input: "()[]{}"
Output: true

Example 3:

Input: "(]"
Output: false

Example 4:

Input: "([)]"
Output: false

Example 5:

Input: "{[]}"
Output: true

Solution in C++:

关键点：如何表示括号及其匹配问题

bool isValid(string s) {
        map<char,int> myMap;
        myMap['('] = 1;
        myMap[')'] = 2;
        myMap['['] = 4;
        myMap[']'] = 5;
        myMap['{'] = 7;
        myMap['}'] = 8;

        vector<char> vec;
        size_t i = 0;

        for (auto c : s)
        {
            if(myMap[c] == 0)
                return false;

            if (!vec.empty()){
                if (myMap[vec.back()] != (myMap[c]-1))
                    vec.push_back(c);
                else
                    vec.pop_back();
            } else{
                vec.push_back(c);
            }

        }

        if (vec.empty())
            return true;
        else
            return false;

    }

21. Merge Two Sorted Lists

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.
Example:

Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4

Solution in C++:

方法一：没有头结点

ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
      ListNode* merge = nullptr;
      ListNode* ret = nullptr;
      if (l1 == nullptr) return l2;
      if (l2 == nullptr) return l1;

      if (l1->val < l2->val){
              merge = l1;
              l1 = l1->next;
          } else{
              merge = l2;
              l2 = l2->next;
          }
           ret = merge;
       while(l1 && l2){
          if (l1->val < l2->val){
              merge->next = l1;
              l1 = l1->next;
          } else{
              merge->next = l2;
              l2 = l2->next;
          }
          merge = merge->next;
      }

      if (l1){
          merge->next = l1;
      } else{
          merge->next = l2;
      }

      return ret;
}

方法二：有头结点

ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
      ListNode *head= new ListNode(-1), *cur = head;
      while (l1 && l2) {
          if (l1->val < l2->val) {
              cur->next = l1;
              l1 = l1->next;
          } else {
              cur->next = l2;
              l2 = l2->next;
          }
          cur = cur->next;
      }
      cur->next = l1 ? l1 : l2;
      return head->next;
}

错误写法（只写函数体内的代码）：

ListNode* merge = nullptr;
ListNode* ret = nullptr;
bool flag = true;
while(l1 && l2){
    if (l1->val <= l2->val){
        merge = l1;
        l1 = l1->next;
    } else{
        merge = l2;
        l2 = l2->next;
    }
    if (flag){
        ret = merge;
        flag = false;
    }
    merge = merge->next;
}

if (l1){
    merge = l1;
} else{
    merge = l2;
}

return ret;

错误分析：此处写法，merge的作用是想将l1和l2链表的结点链接起来，但是实际作用merge指针只是指向了不同的结点，而并未改变结点next指针域的指向。

小结

1.练习了C++中map和vector的使用
2.对链表的知识进行了复习，但是发现掌握的并不好，还有对指针的理解也存在一定的问题。之后会对链表知识专门进行整理，并且在日后的刷题过程中也需要对链表的使用进行总结。
3.链表中结点之间的链接改变是通过next指针域的改变而改变的。

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