单调栈
单调栈：不从头部出的单调队列
总结
单调队列：擅长维护区间【最大/最小】值，最小值对应单调递增队列
单调栈：擅长维护最近【大于/小于】关系
从左侧先入栈，就是维护左侧最近关系
从右侧先入栈，就是维护右侧最近关系
int main(){int n;cin>>n;vector<int> arr(n);vector<int> pre(n), nxt(n);stack<int> s;for(int i = 0, a;i < n; ++i) cin >> arr[i];for(int i = 0; i < n; i++){while(s.size() && arr[i] < arr[s.top()]){nxt[s.top()] = i;s.pop();}if(s.size() == 0) pre[i] = -1;else pre[i] = s.top();s.push(i);}while(s.size()) nxt[s.top()] = n, s.pop();return 0;
}
单调栈基础题目
155. 最小栈
class MinStack {
public:stack<int> s, min_s;MinStack() {}void push(int val) {s.push(val);if(min_s.size() == 0 || val <= min_s.top()){min_s.push(val);}return;}void pop() {if(s.top() == min_s.top()) min_s.pop();s.pop();return;}int top() {return s.top();}int getMin() {return min_s.top();}
};
503. 下一个更大元素 II
class Solution {
public:vector<int> nextGreaterElements(vector<int>& nums) {stack<int> s;vector<int> ret(nums.size());for(int i = 0; i < nums.size(); i++) ret[i] = -1;for(int i = 0; i < nums.size(); i++){while(s.size() && nums[i] > nums[s.top()]){ret[s.top()] = nums[i];s.pop();}s.push(i);}for(int i = 0; i < nums.size(); i++){while(s.size() && nums[i] > nums[s.top()]){ret[s.top()] = nums[i];s.pop();}s.push(i);}return ret;}
};
901.股票价格跨度
class StockSpanner {
public:typedef pair<int, int> PII;int t;stack<PII> s;StockSpanner() {t = 0;s.push(PII(INT_MAX, t++));}int next(int price) {while(s.size() && price >= s.top().first) s.pop();int ret = t - s.top().second;s.push(PII(price, t++));return ret;}
};
739. 每日温度
class Solution {
public:vector<int> dailyTemperatures(vector<int>& temperatures) {vector<int> ret(temperatures.size());stack<int> s;for(int i = 0; i < temperatures.size(); ++i){while(s.size() && temperatures[i] > temperatures[s.top()]){ret[s.top()] = i - s.top();s.pop();}s.push(i);}return ret;}
};
84. 柱状图中最大的矩形
class Solution {
public:int largestRectangleArea(vector<int>& heights) {stack<int> s;vector<int> l(heights.size()), r(heights.size());int n = heights.size();for(int i = 0; i < n; i++) l[i] = -1, r[i] = n;for(int i = 0; i < n; i++) {while(s.size() && heights[i] <= heights[s.top()]){r[s.top()] = i;s.pop();}if(s.size()) l[i] = s.top();s.push(i);}int ans = 0;for(int i = 0; i < n; i++){ans = max(ans, heights[i] * (r[i] - l[i] - 1));}return ans;}
};
1856. 子数组最小乘积的最大值
class Solution {
public:int maxSumMinProduct(vector<int>& nums) {stack<int> s;vector<int> l(nums.size()), r(nums.size());int n = nums.size();for(int i = 0; i < n; i++) l[i] = -1, r[i] = n;for(int i = 0; i < n; i++) {while(s.size() && nums[i] <= nums[s.top()]){r[s.top()] = i;s.pop();}if(s.size()) l[i] = s.top();s.push(i);}vector<long long> sum(n + 1);sum[0] = 0;for(int i = 0; i < n; i++) sum[i + 1] = sum[i] + nums[i];long long ans = 0;for(int i = 0; i < n; i++){ans = max(ans, nums[i] * (sum[r[i]] - sum[l[i] + 1]));}return ans % (long long)(1e9 + 7); }
};
907. 子数组的最小值之和
class Solution {
public:int sumSubarrayMins(vector<int>& arr) {stack<int> s;int mod_num = 1e9 + 7;long long ans = 0;vector<long long> sum(arr.size() + 1);sum[0] = 0;for(int i = 0; i < arr.size(); i++){while(s.size() && arr[i] <= arr[s.top()]) s.pop();int ind = s.size() ? s.top() : -1;s.push(i);sum[s.size()] = (sum[s.size() - 1] + arr[i] * (i - ind)) % mod_num;ans += sum[s.size()];ans %= mod_num;}return ans;}
};
496. 下一个更大元素 I
class Solution {
public:vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) {unordered_map<int, int> h;stack<int> s;for(auto x: nums2){while(s.size() && x > s.top()){h[s.top()] = x;s.pop();}s.push(x);}vector<int> ret(nums1.size());for(int i = 0; i < nums1.size(); i++){if(h.find(nums1[i]) == h.end()) ret[i] = -1;else ret[i] = h[nums1[i]]; }return ret;}
};
456. 132 模式
- 假设当前元素是中间的最大值，那么需要向前寻找小于他的最小值和向后找大于他的最大值，进行比较看是否符合条件
class Solution {
public:bool find132pattern(vector<int>& nums) {vector<int> l(nums.size());l[0] = INT_MAX;for(int i = 1; i < nums.size(); i++) l[i] = min(l[i - 1], nums[i - 1]);stack<int> s;for(int i = nums.size() - 1; i >= 0; --i){int val = nums[i] + 1;while(s.size() && nums[i] > s.top()) val = s.top(), s.pop();s.push(nums[i]);if(l[i] < nums[i] && val < nums[i] && val > l[i]) return true;}return false;}
};
42. 接雨水
class Solution {
public:int trap(vector<int>& height) {int ans = 0;stack<int> s;for(int i = 0; i < height.size(); i++){while(s.size() && height[i] > height[s.top()]){int now = s.top();s.pop();if(s.size() == 0) continue;int a = height[i] - height[now];int b = height[s.top()] - height[now];ans += (i - s.top() - 1) * min(a, b);}s.push(i);}return ans;}
};
方法详解)
