代码随想录leetcode200题之双指针法

YMWM_2024-05-13 21:47

目录

  • [1 介绍](#1 介绍)
  • [2 训练](#2 训练)
  • [3 参考](#3 参考)

1 介绍

本博客用来记录代码随想录leetcode200题中双指针算法部分的题目。

2 训练

题目1 ：27. 移除元素

C++代码如下，

cpp 复制代码 
class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        int j = 0;
        for (int i = 0; i < nums.size(); ++i) {
            if (nums[i] == val) {
                //
            } else {
                nums[j++] = nums[i];
            }
        }
        return j;
    }
};

python3代码如下，

py 复制代码 
class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        j = 0
        for i in range(len(nums)):
            if nums[i] == val:
                pass
            else:
                nums[j] = nums[i]
                j += 1
        return j

题目2 ：344. 反转字符串

C++代码如下，

cpp 复制代码 
class Solution {
public:
    void reverseString(vector<char>& s) {
        int l = 0, r = s.size() - 1;
        while (l <= r) swap(s[l++], s[r--]);
        return;
    }
};

python3代码如下，

py 复制代码 
class Solution:
    def reverseString(self, s: List[str]) -> None:
        """
        Do not return anything, modify s in-place instead.
        """
        l = 0
        r = len(s) - 1
        while l <= r:
            s[l], s[r] = s[r], s[l]
            l += 1
            r -= 1
        return

题目3 ：54. 替换数字（第八期模拟笔试）

C++代码如下，

cpp 复制代码 
#include <iostream>
#include <string>

using namespace std;

int main() {
    string s;
    cin >> s;
    
    string res;
    for (auto c : s) {
        if (c >= 'a' && c <= 'z') res += c;
        else res += "number";
    }
    cout << res << endl;
    
    return 0;
}

python3代码如下，

py 复制代码 
s = str(input())
res = ""
for c in s:
    if c.isdigit():
        res += "number"
    else:
        res += c 
print(res)

题目4 ：151. 反转字符串中的单词

C++代码如下，

cpp 复制代码 
class Solution {
public:
    string reverseWords(string s) {
        vector<string> words;
        int n = s.size();
        for (int i = 0; i < n; ++i) {
            if (s[i] != ' ') {
                string t = "";
                while (i < n && s[i] != ' ') {
                    t += s[i];
                    i += 1;
                }
                words.emplace_back(t);
            }
        }

        reverse(words.begin(), words.end());
        string res = "";
        for (auto word : words) {
            res += word + " ";
        }
        res = res.substr(0, res.size()-1);
        return res;
    }
};

python3代码如下，

py 复制代码 
class Solution:
    def reverseWords(self, s: str) -> str:
        ls = s.split()
        ls = ls[::-1]
        res = ""
        for x in ls:
            res += x + " "
        res = res[0:-1]
        return res

题目5 ：206. 反转链表

C++代码如下，

cpp 复制代码 
/**
 * Definition for singly-linked list.
 * 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* reverseList(ListNode* head) {
        if (head == nullptr || head->next == nullptr) { //特判空集或一个结点的时候
            return head;
        }

        ListNode* a = head;
        ListNode* b = a->next;

        while (a != nullptr && b != nullptr) {
            ListNode* t = b->next;
            b->next = a;

            a = b;
            b = t;
        }
        
        head->next = nullptr;
        return a;
    }
};

python3代码如下，

py 复制代码 
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        if head is None or head.next is None: #特判空集和一个结点的情况
            return head 

        a = head 
        b = a.next 
        while a is not None and b is not None:
            t = b.next

            b.next = a 

            a = b
            b = t 
        
        head.next = None 
        return a

题目6 ：19. 删除链表的倒数第 N 个结点

C++代码如下，

cpp 复制代码 
/**
 * Definition for singly-linked list.
 * 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* removeNthFromEnd(ListNode* head, int k) {
        //删除倒数第k个结点，快慢指针法

        ListNode* dummy = new ListNode(0, head);

        ListNode* a = dummy;
        ListNode* b = dummy;

        while (k--) { //b先走k步
            b = b->next;
        }

        while (b->next != nullptr) { //b走到最后一个结点
            a = a->next;
            b = b->next;
        }

        if (a->next != nullptr) a->next = a->next->next;
        else cout << "a->next is nullptr!" << endl;

        return dummy->next;
    }
};

python3代码如下，

py 复制代码 
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def removeNthFromEnd(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
        dummy = ListNode(0, head)
        a = dummy
        b = dummy 

        while k > 0:
            b = b.next
            k -= 1

        while b.next is not None:
            b = b.next 
            a = a.next 
        
        a.next = a.next.next 
        return dummy.next

题目7 ：面试题 02.07. 链表相交

C++代码如下，

cpp 复制代码 
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        if(headA == nullptr || headB == nullptr) { //特判都为空集，或者有一方为空集的情况
            return nullptr;
        }
        
        ListNode* a = headA;
        ListNode* b = headB;

        bool flaga = true;
        bool flagb = true;
        while (a != b) {
            a = a->next;
            b = b->next;

            if (a == nullptr) {
                if (flaga) {
                    a = headB;
                    flaga = false;
                } else {
                    break;
                }
            }

            if (b == nullptr) {
                if (flagb) {
                    b = headA;
                    flagb = false;
                } else {
                    break;
                }
            }
        }

        if (a == b) return a;
        else return nullptr;
    }
};

python3代码如下，

py 复制代码 
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
        if headA is None or headB is None:
            return None 
        
        a = headA 
        b = headB 
        flaga = True
        flagb = True 
        while a != b:
            a = a.next 
            b = b.next 

            if a is None:
                if flaga:
                    a = headB 
                    flaga = False
                else:
                    break 
            
            if b is None:
                if flagb:
                    b = headA 
                    flagb = False 
                else:
                    break 
        
        if a == b:
            return a 
        else:
            return None

题目8 ：142. 环形链表 II

C++代码如下，

cpp 复制代码 
/**
 * 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) {
        //这次的快慢指针，是速度上的快慢指针，而非提前走的快慢指针
        ListNode* a = head;
        ListNode* b = head;

        while (b != nullptr && b->next != nullptr) {
            a = a->next;
            b = b->next->next;

            //起点到环入口的距离为x
            //快慢指针相遇点到环入口的距离为y
            //环的周长为c
            //有x % c = y恒成立
            if (a == b) { 
                ListNode* t = head;
                while (t != a) {
                    t = t->next;
                    a = a->next;
                }
                return a;
            }
        }
        return nullptr;
    }
};

python3代码如下，

py 复制代码 
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]:
        a = head 
        b = head 

        while b is not None and b.next is not None:
            a = a.next 
            b = b.next.next 

            if a == b:
                t = head 
                while t != a:
                    t = t.next 
                    a = a.next 
                return a 
        return None

题目9 ：15. 三数之和

C++代码如下，

cpp 复制代码 
class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        map<int, int> map_val_cnt;
        for (auto x : nums) map_val_cnt[x]++;

        vector<int> vals;
        for (auto [k, v] : map_val_cnt) vals.emplace_back(k);

        set<int> vis;
        for (auto x : vals) vis.insert(x);

        int n = vals.size();
        set<vector<int>> ans;
        for (int i = 0; i < n; ++i) {
            int a = vals[i];
            for (int j = 0; j < n; ++j) {
                int b = vals[j];
                int c = 0 - a - b;
                if (vis.count(c) != 0) {
                    unordered_map<int,int> curr_cnt;
                    curr_cnt[a]++;
                    curr_cnt[b]++;
                    curr_cnt[c]++;
                    if (curr_cnt[a] <= map_val_cnt[a] && 
                    curr_cnt[b] <= map_val_cnt[b] &&
                    curr_cnt[c] <= map_val_cnt[c]) {
                        vector<int> t = {a, b, c};
                        sort(t.begin(), t.end());
                        ans.insert(t);
                    }
                }
            }
        }

        vector<vector<int>> res(ans.begin(), ans.end());
        return res;
    }
};

python3代码如下，

py 复制代码 
class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        map_val_cnt = collections.defaultdict(int)
        for x in nums:
            map_val_cnt[x] += 1
        
        vals = []
        for val in map_val_cnt:
            vals.append(val)

        vis = set()
        for val in vals:
            vis.add(val)
        
        ans = set()
        n = len(vals)
        for i in range(n):
            a = vals[i]
            for j in range(n):
                b = vals[j]
                c = 0 - a - b
                if c in vis:
                    curr_cnt = collections.defaultdict(int)
                    curr_cnt[a] += 1
                    curr_cnt[b] += 1
                    curr_cnt[c] += 1

                    if curr_cnt[a] <= map_val_cnt[a] and \
                    curr_cnt[b] <= map_val_cnt[b] and \
                    curr_cnt[c] <= map_val_cnt[c]:
                        t = [a, b, c]
                        t.sort()
                        ans.add(tuple(t))
        ans = list(ans)
        ans = [list(x) for x in ans]
        return ans

题目10 ：18. 四数之和

C++代码如下，

cpp 复制代码 
typedef long long LL;

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        map<int, int> map_val_cnt;
        for (auto x : nums) map_val_cnt[x] += 1;

        vector<int> vals;
        set<int> vis;
        for (auto [k, v] : map_val_cnt) {
            vals.emplace_back(k);
            vis.insert(k);
        }

        set<vector<int>> ans;
        int n = vals.size();
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                for (int k = 0; k < n; ++k) {
                    int a = vals[i];
                    int b = vals[j];
                    int c = vals[k];
                    LL t = (LL)target - a - b - c;
                    
                    if (t > INT_MAX || t < INT_MIN) { //如果t超出了整型范围，那么vis.count(d)一定为0
                        continue;
                    }

                    int d = (LL)target - a - b - c;

                    if (vis.count(d) != 0) { 
                        unordered_map<int,int> curr_cnt;
                        curr_cnt[a]++;
                        curr_cnt[b]++;
                        curr_cnt[c]++;
                        curr_cnt[d]++;
                        if (curr_cnt[a] <= map_val_cnt[a] && curr_cnt[b] <= map_val_cnt[b] &&
                        curr_cnt[c] <= map_val_cnt[c] && curr_cnt[d] <= map_val_cnt[d]) {
                            vector<int> t = {a, b, c, d};
                            sort(t.begin(), t.end());
                            ans.insert(t);
                        }
                    }
                }
            }
        }

        vector<vector<int>> res(ans.begin(), ans.end());
        return res;
    }
};

python3代码如下，

py 复制代码 
class Solution:
    def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
        map_val_cnt = collections.defaultdict(int)
        for x in nums:
            map_val_cnt[x] += 1
        
        vals = []
        vis = set()
        for x in map_val_cnt:
            vals.append(x)
            vis.add(x)
        
        ans = set()
        n = len(vals)
        for i in range(n):
            for j in range(n):
                for k in range(n):
                    a = vals[i]
                    b = vals[j]
                    c = vals[k]
                    d = target - a - b - c 
                    if d in vis:
                        curr_cnt = collections.defaultdict(int)
                        curr_cnt[a] += 1
                        curr_cnt[b] += 1
                        curr_cnt[c] += 1
                        curr_cnt[d] += 1
                        if curr_cnt[a] <= map_val_cnt[a] and curr_cnt[b] <= map_val_cnt[b] and \
                        curr_cnt[c] <= map_val_cnt[c] and curr_cnt[d] <= map_val_cnt[d]:
                            t = [a, b, c, d]
                            t.sort()
                            t = tuple(t)
                            ans.add(t)
        res = list(ans)
        res = [list(x) for x in res]
        return res

3 参考

代码随想录官网

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