力扣单调栈算法专题训练

1 专题说明

本博客用来计算力扣上的单调栈题目、解题思路和代码。

单调栈题目记录：

2232866美丽塔II

在这里插入图片描述

2 训练

题目1：2866美丽塔II。

解题思路：先计算出prefix[i]，表示0~i满足递增情况下，0~i上的元素之和最大值。然后计算出suffix[i]，表示i~n-1满足递增情况下，i~n-1上的元素之和最大值。那么以i为峰顶的美丽塔的元素之和的最大值为prefix[i] + suffix[i] - nums[i]，遍历i，获得答案即可。

本质上，还是可以归类为：找到i左边，并且<=nums[i]的元素值。

C++代码如下，

class Solution {
public:long long maximumSumOfHeights(vector<int>& maxHeights) {int n = maxHeights.size();vector<long long> prefix(n, 0); //prefix[i]表示0~i是递增的情况下，0~i的元素之和stack<int> stk;for (int i = 0; i < n; ++i) {while (!stk.empty() && maxHeights[stk.top()] > maxHeights[i]) {stk.pop();}if (stk.empty()) {prefix[i] = (long long)(i + 1) * maxHeights[i];} else {prefix[i] = prefix[stk.top()] + (long long)(i - stk.top()) * maxHeights[i];}stk.push(i);}while (!stk.empty()) {stk.pop();}vector<long long> suffix(n, 0); //suffix[i]表示i~n-1是递减的情况下，i~n-1的元素之和for (int i = n - 1; i >= 0; --i) {while (!stk.empty() && maxHeights[stk.top()] > maxHeights[i]) {stk.pop();}if (stk.empty()) {suffix[i] = (long long)(n - i) * maxHeights[i];} else {suffix[i] = suffix[stk.top()] + (long long)(stk.top() - i) * maxHeights[i];}stk.push(i);}long long res = 0;for (int i = 0; i < n; ++i) {res = max(res, prefix[i] + suffix[i] - maxHeights[i]);}return res;}
};

python3代码如下，

class Solution:def maximumSumOfHeights(self, maxHeights: List[int]) -> int:n = len(maxHeights)prefix = [0 for i in range(n)] #0~i的递增数组的和的最大值stk = []for i in range(n):while len(stk) and maxHeights[stk[-1]] > maxHeights[i]:del stk[-1]if len(stk) == 0:prefix[i] = (i + 1) * maxHeights[i]else:prefix[i] = prefix[stk[-1]] + (i - stk[-1]) * maxHeights[i]stk.append(i)stk.clear()suffix = [0 for i in range(n)] #i~n-1的递减数组的和的最大值for i in range(n-1,-1,-1):while len(stk) and maxHeights[stk[-1]] > maxHeights[i]:del stk[-1]if len(stk) == 0:suffix[i] = (n - i) * maxHeights[i]else:suffix[i] = suffix[stk[-1]] + (stk[-1] - i) * maxHeights[i]stk.append(i)res = 0for i in range(n):#print(f"i = {i}, prefix[i] = {prefix[i]}, suffix[i] = {suffix[i]}.")res = max(res, prefix[i] + suffix[i] - maxHeights[i])return res

题目2：496下一个更大元素I。

解题思路：直接找右边首次大于它的元素即可。

C++代码如下，

class Solution {
public:vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) {unordered_map<int,int> mp; //mp[x]表示nums2中元素x的右边，第一个比它大的元素stack<int> stk;for (int i = nums2.size() - 1; i >= 0; --i) {while (!stk.empty() && stk.top() <= nums2[i]) {stk.pop();}if (!stk.empty()) {mp[nums2[i]] = stk.top();} else {mp[nums2[i]] = -1;}stk.push(nums2[i]);}vector<int> res;for (auto x : nums1) {res.emplace_back(mp[x]);}return res;}
};

python3代码如下，

class Solution:def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:n = len(nums2)mp = collections.defaultdict(int)stk = []for i in range(n - 1, -1, -1):while len(stk) and stk[-1] <= nums2[i]:del stk[-1]if len(stk):mp[nums2[i]] = stk[-1]else:mp[nums2[i]] = -1stk.append(nums2[i])res = []for x in nums1:res.append(mp[x])return res

题目3：503下一个更大元素II。

解题思路：环形问题，扩展两倍原数组即可，接下来就是找右侧首次大于它的元素。

C++代码如下，

class Solution {
public:vector<int> nextGreaterElements(vector<int>& nums) {int n = nums.size();vector<int> a(2 * n, 0);for (int i = 0; i < n; ++i) {a[i] = a[i + n] = nums[i];}vector<int> ans(2 * n, -1);stack<int> stk;for (int i = 2 * n - 1; i >= 0; --i) {while (!stk.empty() && stk.top() <= a[i]) {stk.pop();}if (!stk.empty()) {ans[i] = stk.top();}stk.push(a[i]);}vector<int> res(n, -1);for (int i = 0; i < n; ++i) {res[i] = ans[i];}return res;}
};

python3代码如下，

class Solution:def nextGreaterElements(self, nums: List[int]) -> List[int]:n = len(nums)a = [-1 for i in range(2 * n)]for i in range(n):a[i] = a[i + n] = nums[i]ans = [-1 for i in range(2 * n)]stk = []for i in range(2 * n - 1, -1, -1):while len(stk) and stk[-1] <= a[i]:del stk[-1]if len(stk):ans[i] = stk[-1]stk.append(a[i])res = [-1 for i in range(n)]for i in range(n):res[i] = ans[i]return res