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思路就是利用俩“指针”i和j，每次i遇到空格，j刚好是单词的首字母，此时不仅仅加入，也将j的位置放到i+1的位置上即可，代码如下：

#include <iostream>
using namespace std;int main() 
{string s;string ret;while (getline(cin, s)){for (int i = 0, j = 0; i < s.size(); i++){if (s[i] == ' '){if (s[j] >= 97){s[j] -= 32;}ret.push_back(s[j]);j = i + 1;}if (i == s.size() - 1){if(s[j] >= 97){s[j] -= 32;}ret.push_back(s[j]);}}cout << ret;}return 0;
}

dd爱框框就是一个简单的滑动窗口问题，进窗口，判断，出窗口。

#include <iostream>
#include <vector>
#include <climits>
using namespace std;
const int N = 1e7 + 10;
int n = 0;
int x = 0;
int arr[N];
int ret;
int rightlen = -1;
int leftlen = -1;
int dis = N;
int main()
{cin >> n >> x;for (int i = 1; i <= n; ++i){cin >> arr[i];}// 滑动窗口int left = 0;int right = 0;while (right <= n){ret += arr[right];while (ret >= x){if (right - left + 1 < dis){leftlen = left;rightlen = right;dis = right - left + 1;}ret -= arr[left];left++;}right++;}cout << leftlen << " " << rightlen;return 0;
}

使用dfs这个思路，每次只要是递归层次到了与digits.size()层次的时候，此时就插入先前排列组合过的字母就行了，其他层级就全排列就用的是dfs。

class Solution {
public:string hash[10] = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};vector<string> ret;string path;vector<string> letterCombinations(string digits) {if (!digits.size()) return {};dfs(0, digits);return ret;}void dfs(int pos, string& digits){if (pos == digits.size()){ret.push_back(path);return;}int index = digits[pos] - '0';for (int i = 0; i < hash[index].size(); ++i){path += hash[index][i];dfs(pos + 1, digits);path.pop_back();}}
};

使用栈，左括号配合右括号，配合到了就返回true，其他都返回false。

class Solution {
public:bool isValid(string s) {if (s.size() == 0) return false;stack<char> st;for (int i = 0; i < s.size(); i++){if (s[i] == '[' || s[i] == '(' || s[i] == '{'){st.push(s[i]);}else{if (st.empty())  return false;char top = st.top();switch(top){case '[':{if (s[i] != ']') return false;break;}case '(':{if (s[i] != ')') return false;break;}case '{':{if (s[i] != '}') return false;break;}default: break;}st.pop();}}if (!st.empty()) return false;return true;}
};