lc1372
（DFS）遍历二叉树，分别从根节点开始尝试向左、向右的 zigzag 路径
记录最长路径长度并返回
sum: 题目加条件，代码dfs就可以加参数...
class Solution {
public:
int longestZigZag(TreeNode* root) {
int ans=0;
function<void(TreeNode*,int,int)>dfs=\&(TreeNode* root,int sum,int dire)
{
if(!root){
ans=max(ans,sum);
return ;
}
if(dire==0){
dfs(root->left,sum+1,1);
dfs(root->right,1,0);
}
else{
dfs(root->right,sum+1,0);
dfs(root->left,1,1);
}
};
dfs(root,0,0);
dfs(root,0,1);
return ans-1;
}
};
lc769
class Solution {
public:
int maxChunksToSorted(vector<int>& arr) {
int ans = 0, mx = 0;
for (int i = 0; i < arr.size(); ++i) {
mx = max(mx, arri);
ans += i == mx;
}
return ans;
}
};
lc768
! 单增栈的大小即为答案
class Solution {
public:
int maxChunksToSorted(vector<int>& arr) {
stack<int> stk;
for (int& v : arr) {
if (stk.empty() || stk.top() <= v)
stk.push(v);
else {
int mx = stk.top();
stk.pop();
while (!stk.empty() && stk.top() > v) stk.pop();
stk.push(mx);
}
}
return stk.size();
}
};
lc388
记录各层级目录长度，遍历输入字符串，找出包含后缀的文件的最长路径长度
class Solution {
public:
int lengthLongestPath(string input) {
int n = input.size();
int res = 0;
vector<int> sum(1000);
int p = 0;
while (p < n) {
int level = 0;
while (p < n && inputp == '\t') {
level++;
p++;
}
int q = p;
bool isfile = false;
while (p < n && inputp != '\n') {
if (inputp == '.')
isfile = true;
p++;
}
if (isfile) {
res = max(res, sumlevel + p - q);
} else {
sumlevel + 1 = sumlevel + p - q + 1;
}
p++;
}
return res;
}
};
lc372_递归 分治 super_pow
快速幂_倍增
用快速幂结合递归，计算a的由数组b表示的幂对1337取模的结果
class Solution {
public:
int MOD = 1337;
int qp(int a, int b) {
a %= MOD;
int t = 1;
while(b) {
if (b%2 == 1) {
t *= a;
t %= MOD;
}
a *= a;
a %= MOD;
b >>= 1;
}
return t;
}
int superPow(int a, vector<int>& b) {
if (b.size() == 0) return 1;
int p = b.back();
b.pop_back();
return qp(superPow(a, b), 10) * qp(a, p) % MOD;
}
};
lc2347
hash
class Solution {
public:
string bestHand(vector<int>& ranks, vector<char>& suits)
{
++bool flush = true;++
for (int i = 1; i < 5 && flush; ++i) {
flush = suitsi == suitsi - 1;
}
if (flush) {
return "Flush";
}
int cnt14{};
bool pair = false;
for (int& x : ranks)
{
if (++cntx == 3)
return "Three of a Kind";
pair |= cntx == 2;
}
return pair ? "Pair" : "High Card";
}
};