June24: nextGreaterElement II, use circular [M]

use stack to handle, but len is 2 times of original length
traverse once, so time and space will be both O(N)

daily/June24.cpp

@@ -0,0 +1,56 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+class Solution {
+ public:
+  /**
+   * @brief LC: 503: Next Greater Element II
+   * Use stack
+   * Time: O(N), Space: O(N)
+   *
+   * @param nums
+   * @return vector<int>
+   */
+  vector<int> nextGreaterElements(vector<int>& nums) {
+    int len = nums.size();
+    vector<int> ans(len, -1);
+
+    stack<int> stk;
+    for (int i = 2 * len - 1; i >= 0; i--) {
+      int x = nums[i % len];
+      while (!stk.empty() && x >= stk.top()) {
+        stk.pop();
+      }
+      if (i < len && !stk.empty()) {
+        ans[i] = stk.top();
+      }
+      stk.push(x);
+    }
+    return ans;
+  }
+
+  /**
+   * @brief nextGreaterElementsII, but from left to right
+   * Still Stack
+   * Time: O(N), Space: O(N)
+   *
+   * @param nums
+   * @return vector<int>
+   */
+  vector<int> nextGreaterElementsII(vector<int>& nums) {
+    int len = nums.size();
+    vector<int> ans(len, -1);
+    stack<int> stk;
+    for (int i = 0; i < 2 * len; i++) {
+      int x = nums[i % len];
+      while (!stk.empty() && x > nums[stk.top()]) {
+        ans[stk.top()] = x;
+        stk.pop();
+      }
+      if (i < len) {
+        stk.push(i);
+      }
+    }
+    return ans;
+  }
+};