tags: Tricky,LeetCode101 original: 0, 2, 1, 4, 3, 5, 7, 6 max: 0, 2, 2, 4, 4, 5, 7, 7 sorted: 0, 1, 2, 3, 4, 5, 6, 7 index: 0, 1, 2, 3, 4, 5, 6, 7 As shown above, the position of break point is same to the position of max value of chunks. So here: ...
tags: Algorithm
tags: Monotonic Stack,LeetCode101,LeetCode101: 496. Next Greater Element I Mono-descreasing stack class Solution { public: vector<int> dailyTemperatures(vector<int>& temperatures) { vector<int> res(temperatures.size(), 0); stack<int> st; for (int i = 0; i < temperatures.size(); i++) { while (!st.empty() && temperatures[st.top()] < temperatures[i]) { res[st.top()] = i - st.top(); st.pop(); } st.push(i); } return res; } }; [73,74,75,71,69,72,76,73]
tags: Monotonic Stack,LeetCode101,Binary Search Tree Mono-descreasing stack Key: The largest number is the root, that we can observe in by iteration. We must clear the stack to fill the right side of BST after loop. The last popped element is the left of current node. From top to bottom, the top element is the right side of the element that under the top. class Solution { public: TreeNode* constructMaximumBinaryTree(vector<int>& nums) { vector<TreeNode*> res(nums.size(), nullptr); stack<int> st; // mono-descreasing stack, remove smaller elements before pushing. TreeNode* root = nullptr; for (int i = 0;i < nums.size(); i++) { res[i] = new TreeNode(nums[i]); while (!st.empty() && nums[st.top()] < nums[i]) { int j = st.top(); st.pop(); if (!st.empty() && nums[st.top()] < nums[i]) { res[st.top()]->right = res[j]; } else { res[i]->left = res[j]; } } if (root == nullptr || res[i]->val > root->val) { root = res[i]; } st.push(i); } while (st.size() > 1) { int j = st.top(); st.pop(); res[st.top()]->right = res[j]; } return root; } };
tags: Monotonic Stack,LeetCode101 Mono-increasing stack Key: Some case should move backward as the new value we meeted is larger than it. When we meet 2 in the stack, and here we need move backward. Some case we need move forward, as the following values are the mono-increaing stack: [1, 2, 5, 3, 4] class Solution { public: int findUnsortedSubarray(vector<int>& nums) { stack<int> st; // mono-increasing int left = -1, right = -2; for (int i = 0; i < nums.size(); i++) { while (!st.empty() && nums[st.top()] > nums[i]) { // move backward if (left == -1 || st.top() < left) { left = st.top(); // left should be the previous index } // move forward for (int j = i; j < nums.size() && nums[j] < nums[st.top()]; j++) { if (j > right) { right = j; } } st.pop(); } st.push(i); } return (right - left) + 1; } }; Failed test cases [2,6,4,8,10,9,15] [1, 2, 3, 4] [1] [2,1] [1,3,2,2,2] [1,2,3,3,3] [2,3,3,2,4] [1,2,5,3,4] [1,3,5,2,4]