算法D58 | 单调栈1 | 739. 每日温度 496.下一个更大元素 I

今天正式开始单调栈，这是单调栈一篇扫盲题目，也是经典题。

大家可以读题，思考暴力的解法，然后在看单调栈的解法。 就能感受出单调栈的巧妙

代码随想录
 

Python:

class Solution:def dailyTemperatures(self, temperatures: List[int]) -> List[int]:n = len(temperatures)result = [0] * nstack = [0]for i in range(1, n):if temperatures[i]<=temperatures[stack[-1]]:stack.append(i)else:while len(stack)!=0 and temperatures[i]>temperatures[stack[-1]]:result[stack[-1]] = i - stack[-1]stack.pop()stack.append(i)return result

C++:

class Solution {
public:vector<int> dailyTemperatures(vector<int>& temperatures) {stack<int> stk;vector<int> result(temperatures.size(), 0);stk.push(0);for (int i=1; i<temperatures.size(); i++) {while (!stk.empty() && temperatures[i]>temperatures[stk.top()]) {result[stk.top()] = i - stk.top();stk.pop();}stk.push(i);}return result;}
};

496.下一个更大元素 I 

本题和 739. 每日温度 看似差不多，其实 有加了点难度。

代码随想录

Python:

class Solution:def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:n1, n2 = len(nums1), len(nums2)result = [-1]*n1    stk = [0]for i in range(1, n2):if nums2[i] <= nums2[stk[-1]]:stk.append(i)else:while len(stk)>0 and nums2[i]>nums2[stk[-1]]:if nums2[stk[-1]] in nums1:idx = nums1.index(nums2[stk[-1]])result[idx] = nums2[i]stk.pop()stk.append(i)return result

C++:

class Solution {
public:vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) {vector<int> result(nums1.size(), -1);stack<int> stk;unordered_map<int,int> umap;for (int i=0; i<nums1.size(); i++) umap[nums1[i]] = i;stk.push(0);for (int i=1; i<nums2.size(); i++) {if (nums2[i] <= nums2[stk.top()]) stk.push(i);else {while (!stk.empty() && nums2[i]>nums2[stk.top()]) {if (umap.count(nums2[stk.top()])>0) {int idx = umap[nums2[stk.top()]];result[idx] = nums2[i];}stk.pop();}stk.push(i);}}return result;}
};