1

Author: Ainevsia

notes/src/SUMMARY.md

@@ -182,4 +182,12 @@
 - [day 57](./day57.md)
   - [647. 回文子串](./day56/lc647.md)
   - [516.最长回文子序列](./day56/lc516.md)
+- [day 58](./day58.md)
+  - [739. 每日温度](./day58/lc739.md)
+  - [496.下一个更大元素 I](./day58/lc496.md)
+- [day 59](./day59.md)
+  - [503. 下一个更大元素II](./day58/lc503.md)
+  - [42. 接雨水](./day58/lc42.md)
+- [day 60](./day60.md)
+  - [84. 柱状图中最大的矩形](./day58/lc84.md)
 - [remains](./remains.md)

notes/src/day58.md

@@ -0,0 +1,27 @@
+# 42. 接雨水
+
+需要考虑前两个
+
+```cpp
+class Solution {
+public:
+    int trap(vector<int>& v) {
+        int res = 0;
+        vector<int>s;
+        for(int i=0;i<v.size();i++){
+            while(!s.empty()&&v[s.back()]<=v[i]){
+                int idx = s.back();
+                s.pop_back();
+                if(!s.empty()){
+                    int left =  s.back();
+                    int width = i - left - 1;
+                    int height = min(v[left],v[i]) - v[idx];
+                    res += width * height;
+                }
+            }
+            s.push_back(i);
+        }
+        return res;
+    }
+};
+```

notes/src/day58/lc42.md

@@ -0,0 +1,28 @@
+# 496. 下一个更大元素 I
+
+从代码来看，v1的作用就是用来映射一次，增加了一层映射
+
+```cpp
+class Solution {
+public:
+    vector<int> nextGreaterElement(vector<int>&v1,vector<int>&v2) {
+        vector<int>res(v1.size(),-1);
+        unordered_map<int, int> m; // key:下标元素，value：下标
+        for (int i = 0; i < v1.size(); i++) {
+            m[v1[i]] = i;
+        }
+        vector<int>s;
+        for(int i=0;i<v2.size();i++){
+            while(!s.empty()&&v2[s.back()]<v2[i]){
+                int idx = s.back();
+                s.pop_back();
+                if (m.find(v2[idx])!=m.end()){
+                    res[m[v2[idx]]]=v2[i];
+                }
+            }
+            s.push_back(i);
+        }
+        return res;
+    }
+};
+```

notes/src/day58/lc496.md

@@ -0,0 +1,21 @@
+# 503.下一个更大元素II
+
+```cpp
+class Solution {
+public:
+    vector<int> nextGreaterElements(vector<int>& v) {
+        vector<int>res(v.size(),-1);
+        vector<int>s;
+        int n = v.size();
+        for(int i=0;i<n*2;i++){
+            while(!s.empty()&&v[s.back()]<v[i%n]){
+                int idx = s.back();
+                s.pop_back();
+                res[idx]=v[i%n];
+            }
+            s.push_back(i%n);
+        }
+        return res;
+    }
+};
+```

notes/src/day58/lc503.md

@@ -0,0 +1,24 @@
+# 739. 每日温度
+
+```cpp
+class Solution {
+public:
+    vector<int> dailyTemperatures(vector<int>& v) {
+        vector<int>res(v.size(),0);
+        // 1
+        // 1 1
+        // 1 1 1
+        vector<int>s;
+        for(int i=0;i<v.size();i++){
+            while(!s.empty()&&v[s.back()]<v[i])
+            {
+                int idx=s.back();
+                s.pop_back();
+                res[idx]=i-idx;
+            }
+            s.push_back(i);
+        }
+        return res;
+    }
+};
+```

notes/src/day58/lc739.md

@@ -0,0 +1,36 @@
+# 84.柱状图中最大的矩形
+
+```cpp
+class Solution {
+public:
+    int largestRectangleArea(vector<int>& v) {
+        int res = 0;
+        vector<int>rightfirstsmallerthanmine(v.size(),v.size());
+        vector<int>leftffirstsmallerthanmine(v.size(),-1);
+        vector<int>s;
+        for(int i=0;i<v.size();i++){
+            while(!s.empty()&&v[s.back()]>v[i]){
+                int idx = s.back();
+                s.pop_back();
+                rightfirstsmallerthanmine[idx]=i;
+            }
+            s.push_back(i);
+        }
+        s.clear();
+        for(int i=v.size()-1;i>=0;i--){
+            while(!s.empty()&&v[s.back()]>v[i]){
+                int idx = s.back();
+                s.pop_back();
+                leftffirstsmallerthanmine[idx]=i;
+            }
+            s.push_back(i);
+        }
+        for(int i=0;i<v.size();i++){
+            int width = rightfirstsmallerthanmine[i] - leftffirstsmallerthanmine[i]-1;
+            int height = v[i];
+            res = max(res, width*height);
+        }
+        return res; 
+    }
+};
+```