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# 算法模板class Solution {
public:
int searchInsert(vector<int>& nums, int target) {
int n = nums.size();
int left = 0;
int right = n; // 我们定义target在左闭右开的区间里，[left, right)
while (left < right) { // 因为left == right的时候，在[left, right)是无效的空间
int middle = left + ((right - left) >> 1);
if (nums[middle] > target) {
right = middle; // target 在左区间，因为是左闭右开的区间，nums[middle]一定不是我们的目标值，所以right = middle，在[left, middle)中继续寻找目标值
} else if (nums[middle] < target) {
left = middle + 1; // target 在右区间，在 [middle+1, right)中
} else { // nums[middle] == target
return middle; // 数组中找到目标值的情况，直接返回下标
}
}
return right;
}
};
void kmp(int* next, const string& s){
next[0] = -1;
int j = -1;
for(int i = 1; i < s.size(); i++){
while (j >= 0 && s[i] != s[j + 1]) {
j = next[j];
}
if (s[i] == s[j + 1]) {
j++;
}
next[i] = j;
}
}二叉树的定义：
struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};前序遍历（中左右）
void traversal(TreeNode* cur, vector<int>& vec) {
if (cur == NULL) return;
vec.push_back(cur->val); // 中 ，同时也是处理节点逻辑的地方
traversal(cur->left, vec); // 左
traversal(cur->right, vec); // 右
}中序遍历（左中右）
void traversal(TreeNode* cur, vector<int>& vec) {
if (cur == NULL) return;
traversal(cur->left, vec); // 左
vec.push_back(cur->val); // 中 ，同时也是处理节点逻辑的地方
traversal(cur->right, vec); // 右
}后序遍历（左右中）
void traversal(TreeNode* cur, vector<int>& vec) {
if (cur == NULL) return;
traversal(cur->left, vec); // 左
traversal(cur->right, vec); // 右
vec.push_back(cur->val); // 中 ，同时也是处理节点逻辑的地方
}相关题解：0094.二叉树的中序遍历
前序遍历（中左右）
vector<int> preorderTraversal(TreeNode* root) {
vector<int> result;
stack<TreeNode*> st;
if (root != NULL) st.push(root);
while (!st.empty()) {
TreeNode* node = st.top();
if (node != NULL) {
st.pop();
if (node->right) st.push(node->right); // 右
if (node->left) st.push(node->left); // 左
st.push(node); // 中
st.push(NULL);
} else {
st.pop();
node = st.top();
st.pop();
result.push_back(node->val); // 节点处理逻辑
}
}
return result;
}
中序遍历（左中右）
vector<int> inorderTraversal(TreeNode* root) {
vector<int> result; // 存放中序遍历的元素
stack<TreeNode*> st;
if (root != NULL) st.push(root);
while (!st.empty()) {
TreeNode* node = st.top();
if (node != NULL) {
st.pop();
if (node->right) st.push(node->right); // 右
st.push(node); // 中
st.push(NULL);
if (node->left) st.push(node->left); // 左
} else {
st.pop();
node = st.top();
st.pop();
result.push_back(node->val); // 节点处理逻辑
}
}
return result;
}后序遍历（左右中）
vector<int> postorderTraversal(TreeNode* root) {
vector<int> result;
stack<TreeNode*> st;
if (root != NULL) st.push(root);
while (!st.empty()) {
TreeNode* node = st.top();
if (node != NULL) {
st.pop();
st.push(node); // 中
st.push(NULL);
if (node->right) st.push(node->right); // 右
if (node->left) st.push(node->left); // 左
} else {
st.pop();
node = st.top();
st.pop();
result.push_back(node->val); // 节点处理逻辑
}
}
return result;
}相关题解：0102.二叉树的层序遍历
vector<vector<int>> levelOrder(TreeNode* root) {
queue<TreeNode*> que;
if (root != NULL) que.push(root);
vector<vector<int>> result;
while (!que.empty()) {
int size = que.size();
vector<int> vec;
for (int i = 0; i < size; i++) {// 这里一定要使用固定大小size，不要使用que.size()
TreeNode* node = que.front();
que.pop();
vec.push_back(node->val); // 节点处理的逻辑
if (node->left) que.push(node->left);
if (node->right) que.push(node->right);
}
result.push_back(vec);
}
return result;
}
可以直接解决如下题目：
int getDepth(TreeNode* node) {
if (node == NULL) return 0;
return 1 + max(getDepth(node->left), getDepth(node->right));
}int countNodes(TreeNode* root) {
if (root == NULL) return 0;
return 1 + countNodes(root->left) + countNodes(root->right);
}void backtracking(参数) {
if (终止条件) {
存放结果;
return;
}
for (选择：本层集合中元素（树中节点孩子的数量就是集合的大小）) {
处理节点;
backtracking(路径，选择列表); // 递归
回溯，撤销处理结果
}
}
int n = 1005; // 根据题意而定
int father[1005];
// 并查集初始化
void init() {
for (int i = 0; i < n; ++i) {
father[i] = i;
}
}
// 并查集里寻根的过程
int find(int u) {
return u == father[u] ? u : father[u] = find(father[u]);
}
// 将v->u 这条边加入并查集
void join(int u, int v) {
u = find(u);
v = find(v);
if (u == v) return ;
father[v] = u;
}
// 判断 u 和 v是否找到同一个根
bool same(int u, int v) {
u = find(u);
v = find(v);
return u == v;
}（持续补充ing）
使用左闭右闭区间
var search = function (nums, target) {
let left = 0, right = nums.length - 1;
// 使用左闭右闭区间
while (left <= right) {
let mid = left + Math.floor((right - left)/2);
if (nums[mid] > target) {
right = mid - 1; // 去左面闭区间寻找
} else if (nums[mid] < target) {
left = mid + 1; // 去右面闭区间寻找
} else {
return mid;
}
}
return -1;
};使用左闭右开区间
var search = function (nums, target) {
let left = 0, right = nums.length;
// 使用左闭右开区间 [left, right)
while (left < right) {
let mid = left + Math.floor((right - left)/2);
if (nums[mid] > target) {
right = mid; // 去左面闭区间寻找
} else if (nums[mid] < target) {
left = mid + 1; // 去右面闭区间寻找
} else {
return mid;
}
}
return -1;
};var kmp = function (next, s) {
next[0] = -1;
let j = -1;
for(let i = 1; i < s.length; i++){
while (j >= 0 && s[i] !== s[j + 1]) {
j = next[j];
}
if (s[i] === s[j + 1]) {
j++;
}
next[i] = j;
}
}二叉树节点定义：
function TreeNode (val, left, right) {
this.val = (val === undefined ? 0 : val);
this.left = (left === undefined ? null : left);
this.right = (right === undefined ? null : right);
}前序遍历（中左右）：
var preorder = function (root, list) {
if (root === null) return;
list.push(root.val); // 中
preorder(root.left, list); // 左
preorder(root.right, list); // 右
}中序遍历（左中右）：
var inorder = function (root, list) {
if (root === null) return;
inorder(root.left, list); // 左
list.push(root.val); // 中
inorder(root.right, list); // 右
}后序遍历（左右中）：
var postorder = function (root, list) {
if (root === null) return;
postorder(root.left, list); // 左
postorder(root.right, list); // 右
list.push(root.val); // 中
}前序遍历（中左右）：
var preorderTraversal = function (root) {
let res = [];
if (root === null) return res;
let stack = [root],
cur = null;
while (stack.length) {
cur = stack.pop();
res.push(cur.val);
cur.right && stack.push(cur.right);
cur.left && stack.push(cur.left);
}
return res;
};中序遍历（左中右）：
var inorderTraversal = function (root) {
let res = [];
if (root === null) return res;
let stack = [];
let cur = root;
while (stack.length ！== 0 || cur !== null) {
if (cur !== null) {
stack.push(cur);
cur = cur.left;
} else {
cur = stack.pop();
res.push(cur.val);
cur = cur.right;
}
}
return res;
};后序遍历（左右中）：
var postorderTraversal = function (root) {
let res = [];
if (root === null) return res;
let stack = [root];
let cur = null;
while (stack.length) {
cur = stack.pop();
res.push(cur.val);
cur.left && stack.push(cur.left);
cur.right && stack.push(cur.right);
}
return res.reverse()
};var levelOrder = function (root) {
let res = [];
if (root === null) return res;
let queue = [root];
while (queue.length) {
let n = queue.length;
let temp = [];
for (let i = 0; i < n; i++) {
let node = queue.shift();
temp.push(node.val);
node.left && queue.push(node.left);
node.right && queue.push(node.right);
}
res.push(temp);
}
return res;
};var getDepth = function (node) {
if (node === null) return 0;
return 1 + Math.max(getDepth(node.left), getDepth(node.right));
}var countNodes = function (root) {
if (root === null) return 0;
return 1 + countNodes(root.left) + countNodes(root.right);
}function backtracking(参数) {
if (终止条件) {
存放结果;
return;
}
for (选择：本层集合中元素（树中节点孩子的数量就是集合的大小）) {
处理节点;
backtracking(路径，选择列表); // 递归
回溯，撤销处理结果
}
} let n = 1005; // 根据题意而定
let father = new Array(n).fill(0);
// 并查集初始化
function init () {
for (int i = 0; i < n; ++i) {
father[i] = i;
}
}
// 并查集里寻根的过程
function find (u) {
return u === father[u] ? u : father[u] = find(father[u]);
}
// 将v->u 这条边加入并查集
function join(u, v) {
u = find(u);
v = find(v);
if (u === v) return ;
father[v] = u;
}
// 判断 u 和 v是否找到同一个根
function same(u, v) {
u = find(u);
v = find(v);
return u === v;
}使用左闭右闭区间
var search = function (nums: number[], target: number): number {
let left: number = 0, right: number = nums.length - 1;
// 使用左闭右闭区间
while (left <= right) {
let mid: number = left + Math.floor((right - left)/2);
if (nums[mid] > target) {
right = mid - 1; // 去左面闭区间寻找
} else if (nums[mid] < target) {
left = mid + 1; // 去右面闭区间寻找
} else {
return mid;
}
}
return -1;
};使用左闭右开区间
var search = function (nums: number[], target: number): number {
let left: number = 0, right: number = nums.length;
// 使用左闭右开区间 [left, right)
while (left < right) {
let mid: number = left + Math.floor((right - left)/2);
if (nums[mid] > target) {
right = mid; // 去左面闭区间寻找
} else if (nums[mid] < target) {
left = mid + 1; // 去右面闭区间寻找
} else {
return mid;
}
}
return -1;
};var kmp = function (next: number[], s: number): void {
next[0] = -1;
let j: number = -1;
for(let i: number = 1; i < s.length; i++){
while (j >= 0 && s[i] !== s[j + 1]) {
j = next[j];
}
if (s[i] === s[j + 1]) {
j++;
}
next[i] = j;
}
}二叉树节点定义：
class TreeNode {
val: number
left: TreeNode | null
right: TreeNode | null
constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
this.val = (val===undefined ? 0 : val)
this.left = (left===undefined ? null : left)
this.right = (right===undefined ? null : right)
}
}前序遍历（中左右）：
var preorder = function (root: TreeNode | null, list: number[]): void {
if (root === null) return;
list.push(root.val); // 中
preorder(root.left, list); // 左
preorder(root.right, list); // 右
}中序遍历（左中右）：
var inorder = function (root: TreeNode | null, list: number[]): void {
if (root === null) return;
inorder(root.left, list); // 左
list.push(root.val); // 中
inorder(root.right, list); // 右
}后序遍历（左右中）：
var postorder = function (root: TreeNode | null, list: number[]): void {
if (root === null) return;
postorder(root.left, list); // 左
postorder(root.right, list); // 右
list.push(root.val); // 中
}前序遍历（中左右）：
var preorderTraversal = function (root: TreeNode | null): number[] {
let res: number[] = [];
if (root === null) return res;
let stack: TreeNode[] = [root],
cur: TreeNode | null = null;
while (stack.length) {
cur = stack.pop();
res.push(cur.val);
cur.right && stack.push(cur.right);
cur.left && stack.push(cur.left);
}
return res;
};中序遍历（左中右）：
var inorderTraversal = function (root: TreeNode | null): number[] {
let res: number[] = [];
if (root === null) return res;
let stack: TreeNode[] = [];
let cur: TreeNode | null = root;
while (stack.length ！== 0 || cur !== null) {
if (cur !== null) {
stack.push(cur);
cur = cur.left;
} else {
cur = stack.pop();
res.push(cur.val);
cur = cur.right;
}
}
return res;
};后序遍历（左右中）：
var postorderTraversal = function (root: TreeNode | null): number[] {
let res: number[] = [];
if (root === null) return res;
let stack: TreeNode[] = [root];
let cur: TreeNode | null = null;
while (stack.length) {
cur = stack.pop();
res.push(cur.val);
cur.left && stack.push(cur.left);
cur.right && stack.push(cur.right);
}
return res.reverse()
};var levelOrder = function (root: TreeNode | null): number[] {
let res: number[] = [];
if (root === null) return res;
let queue: TreeNode[] = [root];
while (queue.length) {
let n: number = queue.length;
let temp: number[] = [];
for (let i: number = 0; i < n; i++) {
let node: TreeNode = queue.shift();
temp.push(node.val);
node.left && queue.push(node.left);
node.right && queue.push(node.right);
}
res.push(temp);
}
return res;
};var getDepth = function (node: TreNode | null): number {
if (node === null) return 0;
return 1 + Math.max(getDepth(node.left), getDepth(node.right));
}var countNodes = function (root: TreeNode | null): number {
if (root === null) return 0;
return 1 + countNodes(root.left) + countNodes(root.right);
}function backtracking(参数) {
if (终止条件) {
存放结果;
return;
}
for (选择：本层集合中元素（树中节点孩子的数量就是集合的大小）) {
处理节点;
backtracking(路径，选择列表); // 递归
回溯，撤销处理结果
}
} let n: number = 1005; // 根据题意而定
let father: number[] = new Array(n).fill(0);
// 并查集初始化
function init () {
for (int i: number = 0; i < n; ++i) {
father[i] = i;
}
}
// 并查集里寻根的过程
function find (u: number): number {
return u === father[u] ? u : father[u] = find(father[u]);
}
// 将v->u 这条边加入并查集
function join(u: number, v: number) {
u = find(u);
v = find(v);
if (u === v) return ;
father[v] = u;
}
// 判断 u 和 v是否找到同一个根
function same(u: number, v: number): boolean {
u = find(u);
v = find(v);
return u === v;
}Java：
def binarysearch(nums, target):
low = 0
high = len(nums) - 1
while (low <= high):
mid = (high + low)//2
if (nums[mid] < target):
low = mid + 1
if (nums[mid] > target):
high = mid - 1
if (nums[mid] == target):
return mid
return -1def kmp(self, a, s):
# a: length of the array
# s: string
next = [0]*a
j = 0
next[0] = 0
for i in range(1, len(s)):
while j > 0 and s[j] != s[i]:
j = next[j - 1]
if s[j] == s[i]:
j += 1
next[i] = j
return next Go：
