1LL(1ll) 与 std::to_string()

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x),left(NULL),right(NULL) {}
 * };
 */
class Solution {
public:
    bool isValidBST(TreeNode* root) {
        return dfs(root,INT_MIN,INT_MAX);
    }
 
    bool dfs(TreeNode *root,long long minv,long long maxv){
        if(!root) return true;
        if(root -> val < minv || root -> val > maxv) return false;
 
        return dfs(root -> left,minv,root -> val -1ll) && dfs(root -> right,root -> val + 1ll,maxv);
    }
};

1LL是long long 类型的1，一个int 型的数和一个long long 型的数做运算返回值是long long 类型的，利用这种方式来防止溢出。

std::to_string() 是c++11引入的函数，其使用方法如下：

string to_string (int val);
string to_string (long val);
string to_string (long long val);
string to_string (unsigned val);
string to_string (unsigned long val);
string to_string (unsigned long long val);
string to_string (float val);
string to_string (double val);
string to_string (long double val);

样例如下：

// to_string example
#include <iostream>   // std::cout
#include <string>     // std::string,std::to_string
 
int main ()
{
  std::string pi = "pi is " + std::to_string(3.1415926);
  std::string perfect = std::to_string(1+2+4+7+14) + " is a perfect number";
  std::cout << pi << 'n';
  std::cout << perfect << 'n';
  return 0;
}

Output:
pi is 3.141593 28 is a perfect number
但std::to_string()无法控制生成字符串的格式，意味着你会遇见这样的情况：
std::cout << std::to_string(0.33) << std::endl; 0.330000