企业官网建设流程全解析

lc2423

// try -- every char

一个变量控制不明白 那就再加一个变量🤓👆🏻

++cnt; // restore

class Solution {

public:

bool equalFrequency(string word)

{

unordered_map<char,int> hash;

for(auto& c:word)

hash[c]++;

// try -- every char

for(auto& [ch, cnt] : hash)

{

--cnt;

bool f=true;

int t = 0; //freq

for(auto& [k, v] : hash)

{

if(v == 0) continue;

if(t == 0) t = v; // init

if(v != t)

{

f=false;

break;

}

}

if(f) return true;

++cnt; // restore

}

return false;

}

};

lc549

pii dfs遍历二叉树

记录每个节点“递增连续长度up”和“递减连续长度down”

计算以该节点为中心的最长连续序列长度

最终得到整棵树的最长连续序列

"递"获取每个node pii

"归"的时候return以该节点为中间的最长链

class Solution {

public:

int ans = 0;

int longestConsecutive(TreeNode* root) {

auto dfs = [&](this auto&& dfs, TreeNode* node) -> pair<int, int> {

if (!node) return {0, 0};

auto [lu, ld] = dfs(node->left);

auto [ru, rd] = dfs(node->right);

int u = 1, d = 1;

if (node->left && node->val == node->left->val + 1) u = max(u, lu + 1);

if (node->right && node->val == node->right->val + 1) u = max(u, ru + 1);

//record two path choice mx

if (node->left && node->val == node->left->val - 1) d = max(d, ld + 1);

if (node->right && node->val == node->right->val - 1) d = max(d, rd + 1);

ans = max(ans, u + d - 1);

return {u, d};

};

dfs(root);

return ans;

}

};

lc3641

滑窗和hash天生一对😋

class Solution {
//set size
//max len
public:
int longestSubarray(vector<int>& nums, int k)
{
int n=nums.size();
unordered_map<int,int> hash;
int l=0,r=0,mx=0,rp=0;
while(r<n)
{
if(++hash[nums[r]]==2)
rp++;
r++;
while(rp>k)
{
if(--hash[nums[l]]==1)
rp--;
l++;
}
mx=max(mx,r-l);
}
return mx;
}
};

rust导入hash

use std::collections::HashMap;

impl Solution {
pub fn longest_subarray(nums: Vec<i32>, k: i32) -> i32 {
let n = nums.len();
let mut hash = HashMap::new();
let mut l = 0;
let mut rp = 0;
let mut mx = 0;

for r in 0..n {
let cnt =hash.entry(nums[r]).and_modify(|c| *c += 1).or_insert(1);
if *cnt == 2 {
rp += 1;
}

while rp > k {
let cnt = hash.entry(nums[l]).and_modify(|c| *c -= 1).or_insert(0);
if *cnt == 1 {
rp -= 1;
}
l += 1;
}

mx = mx.max((r - l + 1) as i32);
}

mx
}
}