【LeetCode】Stack

[503] Next Greater Element II [Medium]

给一个循环数组，找到离当前元素最近的比它大的元素。

Input: [1,2,1]
Output: [2,-1,2]
Explanation: The first 1's next greater number is 2; 
The number 2 can't find next greater number; 
The second 1's next greater number needs to search circularly, which is also 2.

我的思路：直接暴力解。

 class Solution {
 public:
     vector<int> nextGreaterElements(vector<int>& nums) {
         vector<int> ans(nums.size(), -);
         for(auto i = ; i < nums.size(); ++i) {
             bool flag = false;
             for(auto j = i + ; j < nums.size(); ++j) {
                 if (nums[j] > nums[i]) {
                     ans[i] = nums[j];
                     flag = true;
                     break;
                 }
             }
             if (flag == false) {
                 for (auto j = ; j < i; ++j) {
                     if (nums[j] > nums[i]) {
                         ans[i] = nums[j];
                         flag = true;
                         break;
                     }
                 }
             }
         }
         return ans;
     }
 };

[150] Evaluate Reverse Polish Notation [Medium]

  ["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9
  ["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6

注意一下弹出的顺序，不要写错了（- 和 / 对于操作数有顺序之分）

 class Solution {
 public:
     int evalRPN(vector<string>& tokens) {
         stack<int> stk;
         int x, y;
         for (auto token : tokens) {
             if (token == "+" || token == "-" || token == "*" || token == "/") {
                 x = stk.top();
                 stk.pop();
                 y = stk.top();
                 stk.pop();
                 if (token == "+") x = x + y;
                 if (token == "-") x = y - x;
                 if (token == "*") x = x * y;
                 if (token == "/") x = y / x;
                 stk.push(x);
             } else {
                 int t = stoi(token);
                 stk.push(t);
             }
         }
         int ans = stk.top();
         return ans;
     }
 };