NewDS

模板/排序/归并排序.cpp

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+#include <vector>
+#include <queue>
+#include <iostream>
+#include <tuple>
+using namespace std;
+
+void merge(vector<int>& nums, int l, int r){
+    int mid = (l + r) / 2;
+    int i = l, j = mid + 1;
+    vector<int> tmp;
+    while(i <= mid && j <= r){
+        if(nums[i] <= nums[j]) tmp.push_back(nums[i++]);
+        else tmp.push_back(nums[j++]);
+    }
+    while(i <= mid) tmp.push_back(nums[i++]);
+    while(j <= r) tmp.push_back(nums[j++]);
+
+    for(int x = l; x <= r; x++){
+        nums[x] = tmp[x - l];
+    }
+}
+
+void mergeSort(vector<int>& nums, int l, int r){
+    //回溯，后序
+    if(l >= r) return;
+    int mid = (l + r) / 2;
+    mergeSort(nums, l, mid);
+    mergeSort(nums, mid + 1, r);
+    //合并可以写在一起，但是分开清晰
+    merge(nums, l, r);
+}
+
+
+int main(){
+
+    return 0;
+}

模板/树/前中建树BFS.cpp

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+class Solution {
+public:
+    /**
+     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
+     *
+     * 求二叉树的右视图
+     * @param preOrder int整型vector 先序遍历
+     * @param inOrder int整型vector 中序遍历
+     * @return int整型vector
+     */
+    TreeNode* build(
+    const vector<int>& pre, int preL, int preR, 
+    const vector<int>& in, int inL, int inR)
+{
+    if (preL > preR || inL > inR) return nullptr;
+
+    TreeNode* root = new TreeNode(pre[preL]);
+
+    int Search = find(in.begin() + inL, in.begin() + inR + 1, pre[preL]) - in.begin();
+    int k = Search - inL;  // 左子树节点数
+
+    root->left = build(
+        pre,
+        preL + 1,
+        preL + k,
+        in,
+        inL,
+        Search - 1
+    );
+
+    root->right = build(
+        pre,
+        preL + k + 1,
+        preR,
+        in,
+        Search + 1,
+        inR
+    );
+
+    return root;
+}
+
+    vector<int> solve(vector<int>& preOrder, vector<int>& inOrder) {
+    TreeNode* root = build(
+        preOrder, 0, preOrder.size() - 1,
+        inOrder, 0, inOrder.size() - 1
+    );
+
+    vector<int> result;
+    if (!root) return result;
+
+    queue<TreeNode*> q;
+    q.push(root);
+
+    while (!q.empty()) {
+        int sz = q.size();   // 当前层节点数
+        for (int i = 0; i < sz; i++) {
+            TreeNode* t = q.front(); q.pop();
+
+            // 当前层最后一个节点
+            if (i == sz - 1) {
+                result.push_back(t->val);
+            }
+
+            if (t->left) q.push(t->left);
+            if (t->right) q.push(t->right);
+        }
+    }
+    return result;
+}
+
+};