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Commitbaf55b4

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‎README.md‎

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*[二叉树：搜索树的最小绝对差](https://mp.weixin.qq.com/s/Hwzml6698uP3qQCC1ctUQQ)
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*[二叉树：我的众数是多少？](https://mp.weixin.qq.com/s/KSAr6OVQIMC-uZ8MEAnGHg)
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*[二叉树：公共祖先问题](https://mp.weixin.qq.com/s/n6Rk3nc_X3TSkhXHrVmBTQ)
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*[本周小结！（二叉树系列四）](https://mp.weixin.qq.com/s/CbdtOTP0N-HIP7DR203tSg)
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*[二叉树：搜索树的公共祖先问题](https://mp.weixin.qq.com/s/Ja9dVw2QhBcg_vV-1fkiCg)
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*[二叉树：搜索树中的插入操作](https://mp.weixin.qq.com/s/lwKkLQcfbCNX2W-5SOeZEA)
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|[0077.组合](https://github.com/youngyangyang04/leetcode/blob/master/problems/0077.组合.md)|回溯|中等|**回溯**|
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|[0078.子集](https://github.com/youngyangyang04/leetcode/blob/master/problems/0078.子集.md)|回溯/数组|中等|**回溯**|
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|[0083.删除排序链表中的重复元素](https://github.com/youngyangyang04/leetcode/blob/master/problems/0083.删除排序链表中的重复元素.md)|链表|简单|**模拟**|
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|[0084.柱状图中最大的矩形](https://github.com/youngyangyang04/leetcode/blob/master/problems/0084.柱状图中最大的矩形.md)|数组|困难|**单调栈**|
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|[0090.子集II](https://github.com/youngyangyang04/leetcode/blob/master/problems/0090.子集II.md)|回溯/数组|中等|**回溯**|
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|[0093.复原IP地址](https://github.com/youngyangyang04/leetcode/blob/master/problems/0093.复原IP地址)|回溯|中等|**回溯**|
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|[0094.二叉树的中序遍历](https://github.com/youngyangyang04/leetcode/blob/master/problems/0094.二叉树的中序遍历.md)|树|中等|**递归****迭代/栈**|

‎problems/0084.柱状图中最大的矩形.md‎

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```
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如上代码并不能通过leetcode，超时了，因为时间复杂度是O(n^2)。
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##思考一下动态规划
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##单调栈
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单调栈的思路还是不容易理解的，
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想清楚从大到小，还是从小到大，
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本题是从栈底到栈头 从小到大，和 接雨水正好反过来。
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```
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class Solution {
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public:
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int largestRectangleArea(vector<int>& heights) {
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stack<int> st;
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heights.insert(heights.begin(), 0); // 数组头部加入元素0
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heights.push_back(0); // 数组尾部加入元素0
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st.push(0);
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int result = 0;
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// 第一个元素已经入栈，从下表1开始
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for (int i = 1; i < heights.size(); i++) {
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// 注意heights[i] 是和heights[st.top()] 比较 ，st.top()是下表
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if (heights[i] > heights[st.top()]) {
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st.push(i);
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} else if (heights[i] == heights[st.top()]) {
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st.pop(); // 这个可以加，可以不加，效果一样，思路不同
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st.push(i);
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} else {
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while (heights[i] < heights[st.top()]) { // 注意是while
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int mid = st.top();
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st.pop();
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int left = st.top();
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int right = i;
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int w = right - left - 1;
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int h = heights[mid];
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result = max(result, w * h);
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}
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st.push(i);
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}
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}
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return result;
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}
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};
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```
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代码精简之后：
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```
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class Solution {
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public:
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int largestRectangleArea(vector<int>& heights) {
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stack<int> st;
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heights.insert(heights.begin(), 0); // 数组头部加入元素0
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heights.push_back(0); // 数组尾部加入元素0
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st.push(0);
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int result = 0;
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for (int i = 1; i < heights.size(); i++) {
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while (heights[i] < heights[st.top()]) {
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int mid = st.top();
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st.pop();
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int w = i - st.top() - 1;
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int h = heights[mid];
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result = max(result, w * h);
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}
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st.push(i);
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}
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return result;
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}
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};
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```

‎problems/0143.重排链表.md‎

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}
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cur = head;
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int i = 1;
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int j = vec.size() - 1;
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int count = 0; // 计数，用来取前面，取后面
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int j = vec.size() - 1; // i j为之前前后的双指针
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int count = 0; // 计数，偶数去后面，奇数取前面
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while (i <= j) {
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if (count % 2 == 0) {
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cur->next = vec[j];
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cur = cur->next;
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count++;
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}
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if (vec.size() % 2 == 0) {
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if (vec.size() % 2 == 0) { // 如果是偶数，还要多处理中间的一个
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cur->next = vec[i];
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cur = cur->next;
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}
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```
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class Solution {
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public:
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void reorderList(ListNode* head) {
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void reorderList(ListNode* head) {
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deque<ListNode*> que;
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ListNode* cur = head;
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if (cur == nullptr) return;
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}
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cur = head;
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int count = 0;
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int count = 0; // 计数，偶数去后面，奇数取前面
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ListNode* node;
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while(que.size()) {
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if (count % 2 == 0) {
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cur->next = node;
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cur = cur->next;
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}
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cur->next = nullptr;
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cur->next = nullptr; // 注意结尾
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}
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};
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```
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```
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class Solution {
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private:
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// 反转链表
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ListNode* reverseList(ListNode* head) {
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ListNode* temp; // 保存cur的下一个节点
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ListNode* cur = head;
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ListNode* cur2 = head2;
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ListNode* cur = head;
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cur1 = cur1->next;
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int count = 0;
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int count = 0; // 偶数取head2的元素，奇数取head1的元素
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while (cur1 && cur2) {
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if (count % 2 == 0) {
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cur->next = cur2;
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count++;
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cur = cur->next;
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}
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if (cur2 != nullptr) {
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if (cur2 != nullptr) { // 处理结尾
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cur->next = cur2;
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}
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if (cur1 != nullptr) {

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