【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(), -1);
        for(auto i = 0; i < nums.size(); ++i) {
            bool flag = false;
            for(auto j = i + 1; j < nums.size(); ++j) {
                if (nums[j] > nums[i]) {
                    ans[i] = nums[j];
                    flag = true;
                    break;
                }
            }
            if (flag == false) {
                for (auto j = 0; 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;
    }
};