LeetCode Weekly Contest 44解题思路
LeetCode Weekly Contest 44解题思路
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LeetCode Weekly Contest 44解题思路
详细代码可以fork下Github上leetcode项目，不定期更新。
赛题
本次周赛主要分为以下4道题：
- Leetcode 653. Two Sum IV - Input is a BST
- Leetcode 654. Maximum Binary Tree
- Leetcode 655. Print Binary Tree
- Leetcode 656. Coin Path
Leetcode 653. Two Sum IV - Input is a BST
Problem:
Given a Binary Search Tree and a target number, return true if there exist two elements in the BST such that their sum is equal to the given target.
Example 1:
Input:undefined 5 / \ 3 6 / \ \ 2 4 7 Target = 9 Output: True
Example 2:
Input:undefined 5 / \ 3 6 / \ \ 2 4 7 Target = 28 Output: False
典型的TWO SUM，可以采用HashSet，或者中序遍历BST采用两指针方法，时间复杂度均为O(n)O(n)
两指针代码如下：
public boolean findTarget(TreeNode root, int k) {
if (root == null) return false;
nodes = new ArrayList<>();
inorder(root);
int lf = 0, rt = nodes.size() - 1;
while (lf < rt){
int sum = nodes.get(lf) + nodes.get(rt);
if (sum < k){
lf ++;
}
else if (sum > k){
rt --;
}
else{
return true;
}
}
return false;
}
List<Integer> nodes;
public void inorder(TreeNode root){
if (root == null) return;
inorder(root.left);
nodes.add(root.val);
inorder(root.right);
}HashSet版：
public boolean findTarget(TreeNode root, int k) {
return find(root, new HashSet<>(), k);
}
public boolean find(TreeNode root, HashSet<Integer> mem, int k){
if (root == null) return false;
if (mem.contains(k - root.val)) return true;
mem.add(root.val);
return find(root.left, mem, k) || find(root.right, mem, k);
}Leetcode 654. Maximum Binary Tree
Problem:
Given an integer array with no duplicates. A maximum tree building on this array is defined as follow:
- The root is the maximum number in the array.
- The left subtree is the maximum tree constructed from left part subarray divided by the maximum number.
- The right subtree is the maximum tree constructed from right part subarray divided by the maximum number.
Construct the maximum tree by the given array and output the root node of this tree.
嘿，这周咋都是树题，还是采用递归，先找到对应的最大val的index，进行划分，递归的求解即可。
代码如下：
public TreeNode constructMaximumBinaryTree(int[] nums) {
return build(nums, 0, nums.length - 1);
}
TreeNode root = null;
public TreeNode build(int[] nums, int i, int j){
if (i > j) return null;
int idx = findMax(nums, i, j);
TreeNode root = new TreeNode(nums[idx]);
root.left = build(nums, i, idx - 1);
root.right = build(nums, idx + 1, j);
return root;
}
public int findMax(int[] nums, int i, int j){
int max = nums[i];
int id = i;
for (int index = i; index <= j; ++index){
if (max < nums[index]){
max = nums[index];
id = index;
}
}
return id;
}Leetcode 655. Print Binary Tree
Problem:
Print a binary tree in an m*n 2D string array following these rules:
- The row number m should be equal to the height of the given binary tree.
- The column number n should always be an odd number.
- The root node’s value (in string format) should be put in the exactly middle of the first row it can be put. The column and the row where the root node belongs will separate the rest space into two parts (left-bottom part and right-bottom part). You should print the left subtree in the left-bottom part and print the right subtree in the right-bottom part. The left-bottom part and the right-bottom part should have the same size. Even if one subtree is none while the other is not, you don’t need to print anything for the none subtree but still need to leave the space as large as that for the other subtree. However, if two subtrees are none, then you don’t need to leave space for both of them.
- Each unused space should contain an empty string “”.
- Print the subtrees following the same rules.
Example 1:
Input: 1 / 2 Output: [“”, “1”, “”, “2”, “”, “”]
Example 2:
Input: 1 / \ 2 5 /undefined 3undefined /undefined 4undefined Output: [“”, “”, “”, “”, “”, “”, “”, “1”, “”, “”, “”, “”, “”, “”, “” “”, “3”, “”, “”, “”, “”, “”, “”, “”, “”, “”, “”, “”, “”, “”]
Note:
- The height of binary tree is in the range of 1, 10.
依旧递归，先计算树的深度，先把List的空填满，接着对于每一个结点都安排在区间的中间即可。
代码如下：
public List<List<String>> printTree(TreeNode root) {
List<List<String>> ans = new ArrayList<>();
if (root == null) return ans;
int layer = layer(root);
int size = (1 << layer) - 1;
for (int i = 0; i < layer; ++i){
ans.add(new ArrayList<>());
for (int j = 0; j < size; ++j){
ans.get(i).add("");
}
}
dfs(ans, root, 0, 0, size);
return ans;
}
public void dfs(List<List<String>> ans, TreeNode root, int layer, int i, int j){
if (root == null) return;
int mid = (i + j) / 2;
ans.get(layer).set(mid, root.val + "");
dfs(ans,root.left, layer + 1, i, mid - 1);
dfs(ans,root.right, layer + 1, mid + 1, j);
}
public int layer(TreeNode root){
if (root == null) return 0;
int lf = layer(root.left) + 1;
int rt = layer(root.right) + 1;
return Math.max(lf, rt);
}Leetcode 656. Coin Path
Problem:
Given an array A (index starts at 1) consisting of N integers: A1, A2, …, AN and an integer B. The integer B denotes that from any place (suppose the index is i) in the array A, you can jump to any one of the place in the array A indexed i+1, i+2, …, i+B if this place can be jumped to. Also, if you step on the index i, you have to pay Ai coins. If Ai is -1, it means you can’t jump to the place indexed i in the array. Now, you start from the place indexed 1 in the array A, and your aim is to reach the place indexed N using the minimum coins. You need to return the path of indexes (starting from 1 to N) in the array you should take to get to the place indexed N using minimum coins. If there are multiple paths with the same cost, return the lexicographically smallest such path. If it’s not possible to reach the place indexed N then you need to return an empty array.
Example 1:
Input: 1,2,4,-1,2, 2 Output: 1,3,5
Exampel 2:
Input: 1,2,4,-1,2, 1 Output: []
Note:
- Path Pa1, Pa2, …, Pan is lexicographically smaller than Pb1, Pb2, …, Pbm, if and only if at the first i where Pai and Pbi differ, Pai < Pbi; when no such i exists, then n < m.
- A1 >= 0. A2, …, AN (if exist) will in the range of -1, 100.
- Length of A is in the range of 1, 1000.
- B is in the range of 1, 100.
一道DP加路径找回，dpi表示抵达第i个位置的最小代价，转移函数如下：
dp[i] = Math.min(dp[i], dp[i - j] + A[i])
其中 i - j < i的所有状态
紧接着判断dp[n-1]的值，是否INF，如果INF表明没有抵达位置n-1的路径。
反之，我们用-dp[n-1]来标注出路径，并且向回找。
搜索所有负权值的路径，构建出“最小路径”代码如下：
static final int INF = 1 << 29;
public List<Integer> cheapestJump(int[] A, int B) {
List<Integer> ans = new ArrayList<>();
int n = A.length;
if (n == 0) return ans;
int[] dp = new int[n];
Arrays.fill(dp, INF);
dp[0] = A[0];
for (int i = 1; i < n; ++i){
if (A[i] == -1) continue;
for (int j = B; j >= 1; --j){
if (i - j >= 0){
dp[i] = Math.min(dp[i], dp[i - j] + A[i]);
}
}
}
if (dp[n - 1] == INF) return ans;
dp[n - 1] = -dp[n -1];
for (int i = n - 2; i >= 0; --i){
for (int j = B; j >= 1; --j){
if (i + j >= n) continue;
if (dp[i] + A[i + j] == -dp[i + j]){
dp[i] = -dp[i];
}
}
}
ans.add(1);
int cur = 0;
int i = 1;
while (i < n){
int min = INF;
int id = -1;
for (int j = i; j < n; ++j){
if (dp[j] <= 0 && Math.abs(dp[j]) == Math.abs(dp[cur]) + A[j]){
if (min > A[j]){
min = A[j];
id = j;
}
}
}
if (id != -1){
ans.add(id + 1);
cur = id;
i = id;
}
i ++;
}
return ans;
}当然根据题目意思，也只会生成一条最优路径，所以上述while循环可以直接简化成：
while (i < n){
// int min = INF;
int id = -1;
for (int j = i; j < n; ++j){
if (dp[j] <= 0 && Math.abs(dp[j]) == Math.abs(dp[cur]) + A[j]){
id = j;
break;
// if (min > A[j]){
// min = A[j];
// id = j;
// }
}
}
if (id != -1){
ans.add(id + 1);
cur = id;
i = id;
}
i ++;
}题目要求我们求解字典序最小的情况，即达到目的地有多条最优路径时，输出最优路径字典序最小。
接下来的做法需要知道这样一个事实，就论此题，当存在多条最优路径时，长度较长的最优路径，字典序较小。
比如：
A = 0 0 0 0 0
1 2 3 4 5
B = 3
那么我们有最优路径：
1 3 5
1 2 4 5
证明的关键在于位置2为什么一定会存在。
假设位置2不存在，这就意味着，如果存在多条最优路径，必然是如下形式：
最优路径1：1 3 5
可能路径2：1 3 4 5
可能路径3：1 4 5
但上述可能路径可以根据性质推得均不可能存在，因为假设2存在，必然有新的最优路径为1 4 5而不需要经过位置3.
假设3存在，自然比1 3 5还要最优，所以只会出现1 2 4 5这种位置，那么字典序自然是较小的。接着我们可以采用逆序DAG的做法，通过一次遍历解决上述最优路径+字典序最小的选择。
代码如下：
static final int INF = 1 << 29;
public List<Integer> cheapestJump(int[] A, int B) {
int n = A.length;
List<Integer> ans = new ArrayList<>();
if (n == 0) return ans;
int[] dp = new int[n];
int[] path = new int[n];
Arrays.fill(dp, INF);
dp[n - 1] = A[n - 1];
for (int i = n - 2; i >= 0; --i){
if (A[i] == -1) continue;
for (int j = 1; j <= B; ++j){
if (i + j >= n || dp[i + j] < 0) continue;
if (dp[i + j] + A[i] < dp[i]){
dp[i] = dp[i + j] + A[i];
path[i] = i + j;
}
}
}
if (dp[0] == INF) return ans;
ans.add(1);
int cur = 0;
while (cur < n - 1){
ans.add(path[cur] + 1);
cur = path[cur];
}
return ans;
}所以问题的关键在于for循环中变量的【顺序】，这里从后往前构建的原因在于方便从1开始输出路径，最优路径前向和后向结果是一样的。而path和j出现的顺序在于输出字典序最小的情况，因为dpi每次更新时，都是【临近】更新，而如果dpi已经达到最优，根据“>”的特性，后续的【i+j】不会被更新到，那么自然地，从1开始的最优路径，一定是长度最长的，而我们证明了，长度最长的一定是字典序最小的。