LeetCode：14、20、21、26

14、最长公共前缀

#include <iostream>
#include <string.h>
#include <vector>
using namespace std;
class Solution {
  public:
    string longestCommonPrefix(vector<string> &strs) {
        string ans;
        int Size = strs.size();
        if (Size == 0) {
            return ans;
        }
        int MinLen = strs.at(0).length();
        for (int i = 0; i < Size; i++) {
            if (strs.at(i).length() < MinLen) {
                MinLen = strs.at(i).length();
            }
        }
        if (MinLen == 0) {
            return ans;
        }
        bool judge = 1;
        for (int i = 0; i < MinLen; i++) {
            char temp = strs.at(0).at(i);
            for (int j = 1; j < Size; j++) {
                if (strs.at(j).at(i) != temp) {
                    judge = 0;
                    break;
                }
            }
            if (judge) {
                ans.push_back(temp);
            } else {
                break;
            }
        }
        return ans;
    }
    //*123/123 cases passed (0 ms)
    //*Your runtime beats 100 % of cpp submissions
    //*Your memory usage beats 76.45 % of cpp submissions (8.9 MB)
};

20、有效的括号

#include <iostream>
#include <stack> //https://blog.csdn.net/lhyer/article/details/47908063
#include <string.h>
using namespace std;
class Solution {
  public:
    bool judge(char ch1, char ch2) {
        if (ch1 == '(' && ch2 == ')') {
            return 1;
        }
        if (ch1 == '[' && ch2 == ']') {
            return 1;
        }
        if (ch1 == '{' && ch2 == '}') {
            return 1;
        }
        return 0;
    }
    bool isValid(string s) {
        int len = s.length();
        if (len == 0) {
            return 1;
        }
        if (len == 1) {
            return 0;
        }
        stack<char> st;
        st.push(s[0]);
        for (int i = 1; i < len; i++) {
            if (st.empty()) {
                st.push(s[i]);
            } else {
                if (judge(st.top(), s[i])) {
                    st.pop();
                } else {
                    st.push(s[i]);
                }
            }
        }
        if (st.empty()) {
            return 1;
        } else {
            return 0;
        }
    }
};
//*91/91 cases passed (0 ms)
//*Your runtime beats 100 % of cpp submissions
//*Your memory usage beats 26.67 % of cpp submissions (6.3 MB)
// TODO:有必要优化一下空间

21、合并两个有序链表

#include <iostream>
#include <stdlib.h>

using namespace std;
struct ListNode {
    int val;
    ListNode *next;
    ListNode() : val(0), next(nullptr) {}
    ListNode(int x) : val(x), next(nullptr) {}
    ListNode(int x, ListNode *next) : val(x), next(next) {}
};

class Solution {
  public:
    ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
        ListNode l3;
        ListNode *p3 = &l3;
        while (l1 && l2) {
            if (l1->val <= l2->val) {
                p3->next = l1;
                l1 = l1->next;
                p3 = p3->next;
            } else {
                p3->next = l2;
                l2 = l2->next;
                p3 = p3->next;
            }
        }
        p3->next = l1 ? l1 : l2;
        return l3.next;
    }
};
// 208/208 cases passed (12 ms)
// Your runtime beats 48.07 % of cpp submissions
// Your memory usage beats 99.03 % of cpp submissions (14.2 MB)

---

写完这道题，更加深刻地明白了链表排序的方法。

以下是学校《数据结构》上面写的排序函数：

void MergeList_L(LinkList &La, LinkList &Lb, LinkList &Lc) {
    LinkList ptr;
    LinkList pa = La->next;
    LinkList pb = Lb->next; // pa、pb无头部的空节点
    LinkList pc = La;  //和上面不同，开头是对空结点的处理
    Lc = La;           // La的头结点作为Lc的头结点
    while (pa && pb) { //需要比较的段
        if (pa->data < pb->data) {
            pc->next = pa;
            pc = pa;       //更新节点
            pa = pa->next; // pa后移
        } else if (pa->data > pb->data) {
            pc->next = pb;
            pc = pb;
            pb = pb->next;
        } else {
            pc->next = pa; //多了一步删除重复数字的操作
            pc = pa;
            pa = pa->next;
            ptr = pb;
            pb = pb->next;
            free(ptr);
        }
    }
    pc->next = pa ? pa : pb; //插入剩余段
    free(Lb);
}

可以看到，在真正的核心代码部分，我在LeetCode上写的如下：

p3->next = l1;
l1 = l1->next;
p3 = p3->next;

书上的：

pc->next = pb;
pc = pb;
pb = pb->next;

我承认两者就是一个意思，但是为什么后者一定要写成pc=pb？这完全曲解了代码的意思，相比之下p3 = p3->next就一目了然，直接表明移动到下一个。

26、删除有序数组中的重复项

#include <iostream>
#include <vector>
using namespace std;
class Solution {
  public:
    int removeDuplicates(vector<int> &nums) {
        int len = nums.size();
        for (int i = 0; i < len - 1; i++) {
            if (nums.at(i) == nums.at(i + 1)) {
                for (int j = i; j < len - 1; j++) {
                    nums.at(j) = nums.at(j + 1);
                }
                i--;
                len--;
            }
        }
        return len;
    }
    //一遍过，可以！
    // 161/161 cases passed (1648 ms)
    //*应该是用了两个循环的问题，时间复杂度有点大了
    // Your runtime beats 5.36 % of cpp submissions
    // Your memory usage beats 57.44 % of cpp submissions (13.3 MB)
    // TODO：优化这个代码
};