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Commit cd47649

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1008-1028-1340-1449-1467-1687 (6)
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import java.util.ArrayDeque;
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import java.util.Deque;
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public class Solution1008 {
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public TreeNode bstFromPreorder(int[] preorder) {
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Deque<TreeNode> st = new ArrayDeque<>();
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TreeNode root = new TreeNode(preorder[0]);
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st.push(root);
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for (int i = 1; i < preorder.length; i++) {
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TreeNode node = new TreeNode(preorder[i]);
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if (preorder[i] < st.peek().val) { // 小于栈顶元素的值,说明应该在栈顶元素的左子树
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st.peek().left = node;
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} else { // 大于栈顶元素的值,我们要找到当前元素的父节点
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TreeNode parent = st.peek();
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// 栈从栈底到栈顶是递减的
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while (!st.isEmpty() && preorder[i] > st.peek().val) {
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parent = st.pop();
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}
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parent.right = node;
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}
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// 节点压栈
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st.push(node);
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}
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return root;
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}
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}
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/*
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1008. 前序遍历构造二叉搜索树
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https://leetcode.cn/problems/construct-binary-search-tree-from-preorder-traversal/description/
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给定一个整数数组,它表示BST(即 二叉搜索树 )的 先序遍历 ,构造树并返回其根。
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保证 对于给定的测试用例,总是有可能找到具有给定需求的二叉搜索树。
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二叉搜索树 是一棵二叉树,其中每个节点, Node.left 的任何后代的值 严格小于 Node.val , Node.right 的任何后代的值 严格大于 Node.val。
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二叉树的 前序遍历 首先显示节点的值,然后遍历Node.left,最后遍历Node.right。
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提示:
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1 <= preorder.length <= 100
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1 <= preorder[i] <= 10^8
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preorder 中的值 互不相同
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单调栈 https://leetcode.cn/problems/construct-binary-search-tree-from-preorder-traversal/solutions/301677/javadai-ma-de-5chong-jie-ti-si-lu-by-sdwwld/
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时间复杂度 O(n)
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*/
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public class Solution1028 {
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private char[] s;
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private int n, i;
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public TreeNode recoverFromPreorder(String traversal) {
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s = traversal.toCharArray();
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n = s.length;
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i = 0;
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return dfs(0);
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}
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private TreeNode dfs(int level) {
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// 从 i 开始构建 level 层的子树
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int x = 0, h = 0;
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int j = i;
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for (; j < n && s[j] == '-'; ++j) ++h;
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if (h < level) return null; // 非当前层节点 直接返回
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for (; j < n && Character.isDigit(s[j]); ++j) x = x * 10 + (s[j] - '0');
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// 是当前层节点
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i = j; // 更新i
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TreeNode node = new TreeNode(x);
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node.left = dfs(level + 1);
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node.right = dfs(level + 1);
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return node;
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}
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}
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/*
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1028. 从先序遍历还原二叉树
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https://leetcode.cn/problems/recover-a-tree-from-preorder-traversal/description/
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我们从二叉树的根节点 root 开始进行深度优先搜索。
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在遍历中的每个节点处,我们输出 D 条短划线(其中 D 是该节点的深度),然后输出该节点的值。(如果节点的深度为 D,则其直接子节点的深度为 D + 1。根节点的深度为 0)。
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如果节点只有一个子节点,那么保证该子节点为左子节点。
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给出遍历输出 S,还原树并返回其根节点 root。
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提示:
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原始树中的节点数介于 1 和 1000 之间。
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每个节点的值介于 1 和 10 ^ 9 之间。
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递归版本 https://leetcode.cn/problems/recover-a-tree-from-preorder-traversal/solutions/292136/cong-xian-xu-bian-li-huan-yuan-er-cha-shu-by-leetc/comments/2391035/
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时间复杂度:O(|traversal|),其中 |traversal| 是字符串 traversal 的长度。
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*/
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import org.junit.jupiter.api.Assertions;
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import org.junit.jupiter.api.Test;
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public class Solution1008Tests {
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private final Solution1008 solution1008 = new Solution1008();
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@Test
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public void example1() {
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int[] preorder = {8, 5, 1, 7, 10, 12};
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TreeNode expected = TreeNode.buildTreeNode("[8,5,10,1,7,null,12]");
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Assertions.assertTrue(TreeNode.assertTreeNodeEquals(expected, solution1008.bstFromPreorder(preorder)));
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}
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@Test
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public void example2() {
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int[] preorder = {1, 3};
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TreeNode expected = TreeNode.buildTreeNode("[1,null,3]");
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Assertions.assertTrue(TreeNode.assertTreeNodeEquals(expected, solution1008.bstFromPreorder(preorder)));
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}
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}
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import org.junit.jupiter.api.Assertions;
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import org.junit.jupiter.api.Test;
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public class Solution1028Tests {
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private final Solution1028 solution1028 = new Solution1028();
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@Test
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public void example1() {
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String traversal = "1-2--3--4-5--6--7";
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TreeNode expected = TreeNode.buildTreeNode("[1,2,5,3,4,6,7]");
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Assertions.assertTrue(TreeNode.assertTreeNodeEquals(expected, solution1028.recoverFromPreorder(traversal)));
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}
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@Test
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public void example2() {
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String traversal = "1-2--3---4-5--6---7";
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TreeNode expected = TreeNode.buildTreeNode("[1,2,5,3,null,6,null,4,null,7]");
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Assertions.assertTrue(TreeNode.assertTreeNodeEquals(expected, solution1028.recoverFromPreorder(traversal)));
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}
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@Test
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public void example3() {
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String traversal = "1-401--349---90--88";
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TreeNode expected = TreeNode.buildTreeNode("[1,401,null,349,88,90]");
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Assertions.assertTrue(TreeNode.assertTreeNodeEquals(expected, solution1028.recoverFromPreorder(traversal)));
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}
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}

‎leetcode/leetcode-13/src/main/java/Solution1278.java

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https://leetcode.cn/problems/palindrome-partitioning/
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132. 分割回文串 II
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https://leetcode.cn/problems/palindrome-partitioning-ii/
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1745. 分割回文串 IV
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https://leetcode.cn/problems/palindrome-partitioning-iv/description/
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*/
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import java.util.ArrayDeque;
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import java.util.ArrayList;
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import java.util.Arrays;
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import java.util.Deque;
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import java.util.List;
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public class Solution1340 {
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public int maxJumps(int[] arr, int d) {
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int n = arr.length;
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// arr.emplace_back(1e9);
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List<Integer> a = new ArrayList<>();
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for (int v : arr) a.add(v);
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a.add((int) 1e9);
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arr = a.stream().mapToInt(Integer::intValue).toArray();
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// dp[i] 存储从下标 i 开始最多能跳几次
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int[] dp = new int[n + 1];
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Arrays.fill(dp, 1);
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Deque<Integer> st = new ArrayDeque<>(); // st是单调不增的栈
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for (int i = 0; i < n + 1; ++i) {
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while (!st.isEmpty() && arr[st.peek()] < arr[i]) {
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List<Integer> tp = new ArrayList<>();
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tp.add(st.pop());
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// arr中所有和当前栈顶相等的值都要一起弹出
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while (!st.isEmpty() && arr[st.peek()] == arr[tp.getFirst()]) {
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tp.add(st.pop());
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}
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for (int j : tp) {
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if (i - j <= d) {
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dp[i] = Math.max(dp[i], dp[j] + 1);
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}
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if (!st.isEmpty() && j - st.peek() <= d) {
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dp[st.peek()] = Math.max(dp[st.peek()], dp[j] + 1);
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}
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}
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}
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st.push(i);
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}
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// return *max_element(dp.begin(), dp.end() - 1);
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return Arrays.stream(dp).limit(n).max().orElseThrow();
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}
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}
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/*
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1340. 跳跃游戏 V
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https://leetcode.cn/problems/jump-game-v/description/
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给你一个整数数组 arr 和一个整数 d 。每一步你可以从下标 i 跳到:
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- i + x ,其中 i + x < arr.length 且 0 < x <= d 。
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- i - x ,其中 i - x >= 0 且 0 < x <= d 。
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除此以外,你从下标 i 跳到下标 j 需要满足:arr[i] > arr[j] 且 arr[i] > arr[k] ,其中下标 k 是所有 i 到 j 之间的数字(更正式的,min(i, j) < k < max(i, j))。
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你可以选择数组的任意下标开始跳跃。请你返回你 最多 可以访问多少个下标。
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请注意,任何时刻你都不能跳到数组的外面。
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提示:
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1 <= arr.length <= 1000
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1 <= arr[i] <= 10^5
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1 <= d <= arr.length
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单调栈 O(n) https://leetcode.cn/problems/jump-game-v/solutions/3059779/dan-diao-zhan-on-by-utopiainbuaa-0k5c/
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相似题目: 907. 子数组的最小值之和
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https://leetcode.cn/problems/sum-of-subarray-minimums/
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*/
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import org.junit.jupiter.api.Assertions;
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import org.junit.jupiter.api.Test;
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public class Solution1340Tests {
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private final Solution1340 solution1340 = new Solution1340();
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@Test
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public void example1() {
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int[] arr = {6, 4, 14, 6, 8, 13, 9, 7, 10, 6, 12};
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int d = 2;
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int expected = 4;
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Assertions.assertEquals(expected, solution1340.maxJumps(arr, d));
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}
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@Test
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public void example2() {
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int[] arr = {3, 3, 3, 3, 3};
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int d = 3;
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int expected = 1;
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Assertions.assertEquals(expected, solution1340.maxJumps(arr, d));
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}
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@Test
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public void example3() {
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int[] arr = {7, 6, 5, 4, 3, 2, 1};
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int d = 1;
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int expected = 7;
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Assertions.assertEquals(expected, solution1340.maxJumps(arr, d));
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}
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@Test
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public void example4() {
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int[] arr = {7, 1, 7, 1, 7, 1};
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int d = 2;
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int expected = 2;
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Assertions.assertEquals(expected, solution1340.maxJumps(arr, d));
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}
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@Test
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public void example5() {
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int[] arr = {66};
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int d = 1;
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int expected = 1;
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Assertions.assertEquals(expected, solution1340.maxJumps(arr, d));
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}
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}
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import java.util.Arrays;
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public class Solution1449 {
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public String largestNumber(int[] cost, int target) {
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int[] f = new int[target + 1];
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Arrays.fill(f, Integer.MIN_VALUE);
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f[0] = 0;
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for (int i = 1; i <= 9; i++) {
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int u = cost[i - 1];
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for (int j = u; j <= target; j++) {
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f[j] = Math.max(f[j], f[j - u] + 1);
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}
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}
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if (f[target] < 0) return "0";
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StringBuilder ans = new StringBuilder();
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for (int i = 9, j = target; i >= 1; i--) {
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int u = cost[i - 1];
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while (j >= u && f[j] == f[j - u] + 1) {
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ans.append(i);
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j -= u;
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}
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}
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return ans.toString();
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}
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}
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/*
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1449. 数位成本和为目标值的最大数字
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https://leetcode.cn/problems/form-largest-integer-with-digits-that-add-up-to-target/description/
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给你一个整数数组 cost 和一个整数 target 。请你返回满足如下规则可以得到的 最大 整数:
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- 给当前结果添加一个数位(i + 1)的成本为 cost[i] (cost 数组下标从 0 开始)。
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- 总成本必须恰好等于 target 。
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- 添加的数位中没有数字 0 。
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由于答案可能会很大,请你以字符串形式返回。
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如果按照上述要求无法得到任何整数,请你返回 "0" 。
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提示:
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cost.length == 9
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1 <= cost[i] <= 5000
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1 <= target <= 5000
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问题转换为:有若干物品,求给定费用的前提下,花光所有费用所能选择的最大价值(物品个数)为多少。
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https://leetcode.cn/problems/form-largest-integer-with-digits-that-add-up-to-target/solutions/824611/gong-shui-san-xie-fen-liang-bu-kao-lu-we-uy4y/
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时间复杂度 O(nt)。
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*/

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