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参考资料

代码随想录 https://programmercarl.com/
LeetCode https://leetcode.cn/

一 数组

1.1 二分查找

• 前提
  有序数组，元素唯一（有重复元素，使用二分查找法返回的元素下标可能不是唯一）
• 思想
  • 定义 target 是在一个在左闭右闭的区间里，也就是[left, right]
  • while (left <= right) 要使用 <= ，因为left == right是有意义的，所以使用 <=
  • if (nums[middle] > target) right 要赋值为 middle - 1，因为当前这个nums[middle]一定不是target，那么接下来要查找的左区间结束下标位置就是 middle - 1
  • 若要通过二分查找确定插入位置，则最后需判断nums[mid] > target，若是则mid为结果，若不是则mid + 1为结果
• 代码

  // https://leetcode.cn/problems/binary-search/description/
  class Solution {public:int search(vector<int>& nums, int target) {int left = 0, right = nums.size() - 1;while (left <= right) {int mid = left + (right - left)/2;if (nums[mid] == target) {return mid;} if (nums[mid] < target) {left = mid + 1;}if (nums[mid] > target) {right = mid - 1;}}return -1;}
  };
  

1.2 移除元素

• 前提
  O(1)空间，不要求保留原顺序
• 思想
  将要删除的元素用数组最后面的元素进行替换
• 代码

  // https://leetcode.cn/problems/remove-element/description/
  class Solution {public:int removeElement(vector<int>& nums, int val) {int left = 0, right = nums.size() - 1;while (left <= right) {if (nums[left] == val) {nums[left] = nums[right];right --;} else {left ++;}}// [0, right]为需要保留的元素，长度位right + 1return right + 1;}
  };
  

1.3 长度最小的子数组

• 前提
  O(1)空间，子数组连续
• 思想
  定义区间[left, right)，当区间和sum < target时移动right，相反则记录当前区间长度right - left并移动left，直到right > 数组总长度时结束循环
• 代码

  // https://leetcode.cn/problems/minimum-size-subarray-sum/description/
  class Solution {public:int minSubArrayLen(int target, vector<int>& nums) {int left = 0, right = 0; // [left, right)int sum = 0, ans = 0;while (right <= nums.size()) {if (sum < target) {// 此时right指针已达到末尾，且子数组和小于target// 结束循环if (right ==  nums.size()) {break;}sum += nums[right];right ++;} else {int len = right - left;if (ans == 0) {ans = len;} else {ans = ans > len ? len : ans;}sum -= nums[left];left ++;}}return ans;}
  };
  

1.4 螺旋矩阵

• 思想
  关键在于方向的转换，并且需要记录行和列上还有多少是没有遍历过的，每次方向转换都需要减少一个待遍历的行或列
• 代码

  // https://leetcode.cn/problems/spiral-matrix-ii/description/
  class Solution {
  public:vector<vector<int>> generateMatrix(int n) {// direction = 0 => 从左到右，direction = 1 => 从上到下// direction = 2 => 从右到左，direction = 3 => 从下到上int direction = 0;int row_count = n, col_count = n;int i_index = 0, j_index = 0; int num = 1;// 初始化结果vector<vector<int>> res;for (int i = 0; i < n; i ++) {vector<int> r(n);res.push_back(r);}while (num <= n * n) {// printf("direction = %d\n", direction);switch(direction) {case 0:// printf("case0, i = %d, j = %d\n", i_index, j_index);for (int i = 0; i < col_count; i ++) {res[i_index][j_index] = num;j_index ++;num ++;}i_index ++;j_index --;row_count --;break;case 1:// printf("case1, i = %d, j = %d\n", i_index, j_index);for (int i = 0; i < row_count; i ++) {res[i_index][j_index] = num;i_index ++;num ++;}i_index --;j_index --;col_count --;break;case 2:// printf("case2, i = %d, j = %d\n", i_index, j_index);for (int i = 0; i < col_count; i ++) {res[i_index][j_index] = num;j_index --;num ++;}j_index ++;i_index --;row_count --;break;case 3:// printf("case3, i = %d, j = %d\n", i_index, j_index);for (int i = 0; i < row_count; i ++) {res[i_index][j_index] = num;i_index --;num ++;}i_index ++;j_index ++;col_count --;break;}direction ++;direction = direction % 4;// for (int i = 0; i < n; i ++) {//     for (int j = 0; j < n; j ++) {//         cout << res[i][j];//     }//     cout << endl;// }}return res;}
  };
  

1.5 在排序数组中查找元素的第一个和最后一个位置

• 代码

  // https://leetcode.cn/problems/find-first-and-last-position-of-element-in-sorted-array/
  class Solution {
  public:vector<int> searchRange(vector<int>& nums, int target) {vector<int> ret(2, -1);// 有效区间[left, right]int left = 0, right = nums.size() - 1, mid;bool tag = false;while (left <= right) {mid = left + (right - left)/2;if (nums[mid] == target) {tag = true;break;} else if (nums[mid] < target) {left = mid + 1;} else {right = mid - 1;}}if (!tag) return ret;// cout << left << " " <<  mid << " " << right  << endl;// [left, mid - 1] 中寻找小于target的元素int l = left, r = mid - 1, from = left;while (l <= r) {int m = l + (r - l)/2;if (nums[m] < target) {from = m;break;} else {r = m - 1;}}// 向右找到第一个等于target的元素while (nums[from] != target) from ++;// [mid + 1, right] 中寻找大于target的元素l = mid + 1;r = right;int to = right;while (l <= r) {int m = l + (r - l)/2;if (nums[m] > target) {to = m;break;} else {l = m + 1;}}// 向左找到第一个等于target的元素while (nums[to] != target) to --;ret[0] = from;ret[1] = to;return ret;}
  };
  

二 链表

2.1 移除链表元素

• 思想
  移除链表中指定元素的关键在于找到被移除元素的上一个节点
• 代码

  // https://leetcode.cn/problems/remove-element/description/
  class Solution {
  public:ListNode* removeElements(ListNode* head, int val) {// 定义虚拟头节点，不需要对第一个节点进行特殊判断ListNode *dummy_head = new ListNode(-1, head);// 定义pre和cur，方便删除元素ListNode *pre = dummy_head;ListNode *cur = dummy_head->next;while (cur != NULL) {if (cur->val == val) {pre->next = cur->next;cur = pre->next;} else {pre = pre->next;cur = cur->next;}}return dummy_head->next;}
  };
  

2.2 设计链表

• 思想
  • 设计虚拟头节点可以简化对于第一个节点的操作
  • 由题可知，大部分操作都需要找到下标的为index的节点的上一个节点，因此设计了getPre(int index)函数
  • 设计尾节点可以降低addAtTail的时间复杂度，但是在链表变化时需要对其进行维护
• 代码

  // https://leetcode.cn/problems/design-linked-list/
  class MyListNode {    
  public:int val;MyListNode *next;MyListNode(int val, MyListNode *next) {this->val = val;this->next = next;}
  };class MyLinkedList {
  private:MyListNode *dummy_head;MyListNode *tail;int size;// 找到下标为index的上一个节点MyListNode *getPre(int index) {MyListNode *node = dummy_head;while (index > 0) {node = node->next;index --;}return node;}
  public:MyLinkedList() {dummy_head = new MyListNode(-1, NULL);tail = NULL;size = 0;}int get(int index) {if (index >= size) {return -1;}MyListNode *node = getPre(index);return node->next->val;}void addAtHead(int val) {MyListNode *node = new MyListNode(val, NULL);node->next = dummy_head->next;dummy_head->next = node;if (tail == NULL) {tail = node;}size ++;}void addAtTail(int val) {if (tail == NULL) {addAtHead(val);   } else {tail->next = new MyListNode(val, NULL);tail = tail->next;size ++;}}void addAtIndex(int index, int val) {// 如果 index 比长度更大，该节点将 不会插入 到链表中if (index > size) {return ;}// 如果 index 等于链表的长度，那么该节点会被追加到链表的末尾if (index == size) {addAtTail(val);} else {MyListNode *node = new MyListNode(val, NULL);MyListNode *p = getPre(index);node->next = p->next;p->next = node;size ++;}}void deleteAtIndex(int index) {if (index >= size) {return ;}MyListNode *p = getPre(index);// 要删除的节点为尾节点，需要修改tailif (p->next == tail) {p->next = NULL;tail = p;} else {p->next = p->next->next;                }size --;}
  };
  

2.3 反转链表

• 思想
  • 为原链表添加虚拟头节点，删除节点时不需要维护头节点，简化操作（新链表的虚拟头节点同理）
  • 反转链表主要分为节点断开和节点接入，分为两部分实现，可避免思路不清晰，导致链表成环（q->next = NULL可删除）
• 代码

  // https://leetcode.cn/problems/reverse-linked-list/
  class Solution {
  public:ListNode* reverseList(ListNode* head) {// 为原链表添加虚拟头节点，方便节点的删除ListNode *dummy_head = new ListNode(-1, head);ListNode *p = dummy_head;// 为新链表添加虚拟头节点ListNode *new_head = new ListNode(-1);while (p->next != NULL) {// 把节点从原链表断开ListNode *q = p->next;p->next = q->next;q->next = NULL;// 把节点接入到新链表的头部q->next = new_head->next;new_head->next = q;}return new_head->next;}
  };
  

2.4 两两交换链表中的节点

• 思想
  • 多定义变量可一定程度上简化操作（变量不用🪙）
• 代码

  // https://leetcode.cn/problems/swap-nodes-in-pairs/description/
  class Solution {
  public:ListNode* swapPairs(ListNode* head) {ListNode *dummy_head = new ListNode(-1, head);ListNode *p = dummy_head;while (p->next != NULL && p->next->next != NULL) {// a和b为要进行交换的节点，c为后面的节点ListNode *a = p->next;ListNode *b = a->next;ListNode *c = b->next;// 将a和b从原链表中断开（为了方便理解，可省略）p->next = NULL;a->next = NULL;b->next = NULL;// 重新拼接p->next = b;b->next = a;a->next = c;p = a;}   return dummy_head->next;}
  };
  

2.5 删除链表的倒数第N个节点

• 思想
  先让fast跑n步，然后slow和fast再一起跑，fast到达末尾时，slow刚好为倒数第n+1个节点
• 代码

  // https://leetcode.cn/problems/remove-nth-node-from-end-of-list/description/
  class Solution {
  public:ListNode* removeNthFromEnd(ListNode* head, int n) {ListNode *dummy_head = new ListNode(-1, head);ListNode *fast = dummy_head;ListNode *slow = dummy_head;while (n > 0) {fast = fast->next;n --;}while (fast->next != NULL) {fast = fast->next;slow = slow->next;}slow->next = slow->next->next;return dummy_head->next;   }
  };
  

2.6 链表相交

• 代码

  // https://leetcode.cn/problems/intersection-of-two-linked-lists-lcci/
  class Solution {
  public:ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {// 计算链表A和B的节点数int a_count = 0;int b_count = 0;ListNode *p = headA;while (p != NULL) {a_count ++;p = p->next;}p = headB;while (p != NULL) {b_count ++;p = p->next;}// 若两个链表相差sub个节点，则让长的链表先跑sub步while (a_count > b_count) {headA  = headA->next;a_count --;}while (b_count > a_count) {headB = headB->next;b_count --;}// 两个链表同时往前，此时遇到的第一个相同节点即为结果while (headA != headB) {headA = headA->next;headB = headB->next;}return headA;}
  };
  

2.7 环形链表II

• 思想
  • 通过快慢指针法确定链表是否存在环，并找到环中的任意一个节点
  • 遍历环，直至将环中的所有节点添加到set集合中
  • 从head开始遍历，当节点存在与set集合中时，即为环的入口节点
• 代码

  // https://leetcode.cn/problems/linked-list-cycle-ii/
  class Solution {
  public:ListNode *detectCycle(ListNode *head) {// 通过快慢指针法确定链表是否存在环，并找到环中的任意一个节点ListNode *fast = head;ListNode *slow = head;bool tag = false;while (fast != NULL && fast->next != NULL) {fast = fast->next->next;slow = slow->next;if (fast == slow) {tag = true;break;}}if (tag) {// 此时fast和slow都是环中的节点，遍历环，直至将环中的所有节点添加到`set`集合中set<ListNode*> st;while (st.find(fast) == st.end()) {st.insert(fast);fast = fast->next;}// 从`head`开始遍历，当节点存在与`set`集合中时，即为环的入口节点while (st.find(head) == st.end()) {head = head->next;}return head;} else {return NULL;}}
  };
  

• 最优解法

  // https://leetcode.cn/problems/linked-list-cycle-ii/solutions/12616/linked-list-cycle-ii-kuai-man-zhi-zhen-shuang-zhi-/
  class Solution {
  public:ListNode *detectCycle(ListNode *head) {ListNode *fast = head;ListNode *slow = head;bool tag = false;while (fast != NULL && fast->next != NULL) {fast = fast->next->next;slow = slow->next;if (fast == slow) {tag = true;break;}}if (!tag) return NULL;fast = head;while (fast != slow) {fast = fast->next;slow = slow->next;}return fast;}
  };
  

三 哈希表

3.1 有效的字母异位词

• 思想
  • 两个字符串长度不同，直接返回false
  • 遍历字符串s，使用map统计第一个字符串s中每各字符出现的次数
  • 遍历字符串t，若字符不存在于map中，则返回false；存在则进行减操作，当字符次数减为0时，从map移除
  • map为空则表示两个字符串是字母异位词
  • （进阶）由于题目中说明字符串都由小写字母组成，那么我们可以将所有字符映射到长度为26的数组，简化操作
• 代码（基础解法）

  // https://leetcode.cn/problems/valid-anagram/description/
  class Solution {
  public:bool isAnagram(string s, string t) {if (s.length() != t.length()) {return false;}map<char, int> mp;for (int i = 0; i < s.length(); i ++) {char key = s[i];auto it = mp.find(key);if (it != mp.end()) {it->second += 1;} else {mp.insert(pair<char, int>(key, 1));}}for (int i = 0; i < t.length(); i ++) {char key = t[i];auto it = mp.find(key);if (it == mp.end()) {return false;} else {if (it->second == 0) {return false;} else if (it->second == 1) {mp.erase(it);} else {it->second -= 1;}}}return mp.empty();}
  };
  

• 最优解法

  // https://leetcode.cn/problems/valid-anagram/description/
  class Solution {
  public:bool isAnagram(string s, string t) {if (s.length() != t.length()) {return false;}vector<int> mp(26, 0);for (char c : s) {mp[c - 'a'] ++;}for (char c : t) {mp[c - 'a'] --;}for (int c : mp) {if (c != 0) return false;}return true;}
  };
  

3.2 两个数组的交集

• 代码

  // https://leetcode.cn/problems/intersection-of-two-arrays/
  class Solution {
  public:vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {vector<int> ret;set<int> st;// 将nums1的所有元素添加到set中，会自动去重for (int num : nums1) {st.insert(num);}// 遍历nums2，判断元素是否存在st中，存在则是两个数组的交集，添加到结果数组ret中// 为防止元素重复添加，需要将其从st中移除for (int num : nums2) {if (st.find(num) != st.end()) {ret.push_back(num);st.erase(num);}}return ret;}
  };