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Create 2467.Most-Profitable-Path-in-a-Tree.cpp
wisdompeak 168eff5
Update Readme.md
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Create 2471.Minimum-Number-of-Operations-to-Sort-a-Binary-Tree-by-Lev...
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Update Readme.md
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Update 2471.Minimum-Number-of-Operations-to-Sort-a-Binary-Tree-by-Lev...
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50 changes: 50 additions & 0 deletions
...Binary-Tree-by-Level/2471.Minimum-Number-of-Operations-to-Sort-a-Binary-Tree-by-Level.cpp
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,50 @@ | ||
| /** | ||
| * Definition for a binary tree node. | ||
| * struct TreeNode { | ||
| * int val; | ||
| * TreeNode *left; | ||
| * TreeNode *right; | ||
| * TreeNode() : val(0), left(nullptr), right(nullptr) {} | ||
| * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} | ||
| * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} | ||
| * }; | ||
| */ | ||
| class Solution { | ||
| vector<int>level[100005]; | ||
| int maxDepth = 0; | ||
| public: | ||
| int minimumOperations(TreeNode* root) | ||
| { | ||
| dfs(root, 0); | ||
| int count = 0; | ||
| for (int t=0; t<=maxDepth; t++) | ||
| { | ||
| auto& nums = level[t]; | ||
| auto sorted = nums; | ||
| sort(sorted.begin(), sorted.end()); | ||
| unordered_map<int,int>rank; | ||
| for (int i=0; i<sorted.size(); i++) | ||
| rank[sorted[i]] = i; | ||
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| int n = nums.size(); | ||
| for (int i=0; i<n; i++) | ||
| { | ||
| while (rank[nums[i]] != i) | ||
| { | ||
| swap(nums[i], level[t][rank[nums[i]]]); | ||
| count++; | ||
| } | ||
| } | ||
| } | ||
| return count; | ||
| } | ||
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| void dfs(TreeNode* node, int depth) | ||
| { | ||
| if (node==NULL) return; | ||
| maxDepth = max(maxDepth, depth); | ||
| level[depth].push_back(node->val); | ||
| dfs(node->left, depth+1); | ||
| dfs(node->right, depth+1); | ||
| } | ||
| }; |
7 changes: 7 additions & 0 deletions
Greedy/2471.Minimum-Number-of-Operations-to-Sort-a-Binary-Tree-by-Level/Readme.md
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,7 @@ | ||
| ### 2471.Minimum-Number-of-Operations-to-Sort-a-Binary-Tree-by-Level | ||
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| 将属于同一个level的数字都收集起来。然后用Indexing sort的方法去贪心地交换。 | ||
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| 比如说,对于一个乱序的nums数组,我们可以提前知道每个数字的期望位置rank[nums[i]]。我们就从前往后查看每一个位置,如果当前的`rank[nums[i]]!=i`,那么就把nums[i]与位于rank[nums[i]]的数字交换。这样的交换可以持续多次,直至我们在i这个位置上迎来期望的数字。 | ||
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| 为什么一定能够迎来期望的数字呢?因为每一次交换,我们都把一个数字送到了它应该在的位置。一旦把n-1个数字都放到了它们对应期望的位置,那么i这个位置的数字一定也已经安排到了期望的数字。 |
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68 changes: 68 additions & 0 deletions
Tree/2467.Most-Profitable-Path-in-a-Tree/2467.Most-Profitable-Path-in-a-Tree.cpp
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,68 @@ | ||
| class Solution { | ||
| int b[100005]; | ||
| vector<int>next[100005]; | ||
| int ret = INT_MIN/2; | ||
| int bob; | ||
| vector<int>amount; | ||
| public: | ||
| int mostProfitablePath(vector<vector<int>>& edges, int bob, vector<int>& amount) | ||
| { | ||
| this->bob = bob; | ||
| this->amount = amount; | ||
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| int n = amount.size(); | ||
| for (int i=0; i<n; i++) | ||
| b[i] = INT_MAX/2; | ||
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| for (auto& edge: edges) | ||
| { | ||
| int a = edge[0], b = edge[1]; | ||
| next[a].push_back(b); | ||
| next[b].push_back(a); | ||
| } | ||
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| dfs(0, -1, 0); | ||
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| dfs2(0, -1, 0, 0); | ||
| return ret; | ||
| } | ||
|
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| void dfs(int cur, int parent, int step) | ||
| { | ||
| if (cur==bob) | ||
| { | ||
| b[cur] = 0; | ||
| return; | ||
| } | ||
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| int toBob = INT_MAX/2; | ||
| for (int nxt: next[cur]) | ||
| { | ||
| if (nxt==parent) continue; | ||
| dfs(nxt, cur, step+1); | ||
| toBob = min(toBob, b[nxt]+1); | ||
| } | ||
| b[cur] = toBob; | ||
| return; | ||
| } | ||
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| void dfs2(int cur, int parent, int step, int score) | ||
| { | ||
| if (step == b[cur]) | ||
| score += amount[cur]/2; | ||
| else if (step < b[cur]) | ||
| score += amount[cur]; | ||
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| if (next[cur].size()==1 && next[cur][0]==parent) | ||
| { | ||
| ret = max(ret, score); | ||
| return; | ||
| } | ||
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| for (int nxt: next[cur]) | ||
| { | ||
| if (nxt==parent) continue; | ||
| dfs2(nxt, cur, step+1, score); | ||
| } | ||
| } | ||
| }; |
7 changes: 7 additions & 0 deletions
Tree/2467.Most-Profitable-Path-in-a-Tree/Readme.md
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,7 @@ | ||
| ### 2467.Most-Profitable-Path-in-a-Tree | ||
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| 两次遍历全树。第一次遍历求得每个点到bob所在节点的距离(但只限bob节点往上的部分)。第二次遍历求每个点与root的距离。 | ||
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| 对于每个节点而言,如果前者大于后者,则对于Alice而言没有收益。如果前者小于后者,则Alice可以拿取该节点的价值。基于这个原则,第二次dfs的时候可以求得收益最大的一条从root到leaf的路径。 | ||
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| 通过本题,需要掌握不需要建rooted tree的dfs方法,即`dfs(cur, parent)`。当`next[cur]!=parent`时,可以继续递归`dfs(next[cur], cur)`。 |
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