- * 给定两个有序整数数组 nums1 和 nums2,将 nums2 合并到 nums1 中,使得 num1 成为一个有序数组。 - *
- * 说明: - *
- * 初始化 nums1 和 nums2 的元素数量分别为 m 和 n。 - * 你可以假设 nums1 有足够的空间(空间大小大于或等于 m + n)来保存 nums2 中的元素。 - * - * @author yangyi 2019年02月09日11:10:35 - */ -public class MergeArray { - - public int[] merge(int[] nums1, int m, int[] nums2, int n) { - int p = m-- + n-- - 1; - while (m>= 0 && n>= 0) { - nums1[p--] = nums1[m]> nums2[n] ? nums1[m--] : nums2[n--]; - } - - while (n>= 0) { - nums1[p--] = nums2[n--]; - } - return nums1; - } - -} diff --git a/src/main/java/ds/PivotIndex.java b/src/main/java/ds/PivotIndex.java deleted file mode 100644 index 792dbcf..0000000 --- a/src/main/java/ds/PivotIndex.java +++ /dev/null @@ -1,70 +0,0 @@ -package ds; - -import java.util.Arrays; - -/** - * 寻找数组的中心索引 - *
- *
- * 给定一个整数类型的数组 nums,请编写一个能够返回数组"中心索引"的方法。 - *
- * 我们是这样定义数组中心索引的:数组中心索引的左侧所有元素相加的和等于右侧所有元素相加的和。 - *
- * 如果数组不存在中心索引,那么我们应该返回 -1。如果数组有多个中心索引,那么我们应该返回最靠近左边的那一个。 - *
- * 示例 1: - *
- * 输入: - * nums = [1, 7, 3, 6, 5, 6] - * 输出: 3 - * 解释: - * 索引3 (nums[3] = 6) 的左侧数之和(1 + 7 + 3 = 11),与右侧数之和(5 + 6 = 11)相等。 - * 同时, 3 也是第一个符合要求的中心索引。 - * 示例 2: - *
- * 输入: - * nums = [1, 2, 3] - * 输出: -1 - * 解释: - * 数组中不存在满足此条件的中心索引。 - * 说明: - *
- * nums 的长度范围为 [0, 10000]。 - * 任何一个 nums[i] 将会是一个范围在 [-1000, 1000]的整数。 - * - * @author yangyi 2019年02月16日23:46:58 - */ -public class PivotIndex { - - public int pivotIndex(int[] nums) { - if (nums.length < 3) { - return -1; - } - int sum = 0; - int leftSum = 0; - int rightSum = 0; - for (int i = 0; i < nums.length; i++) { - sum += nums[i]; - } - for (int i = 0; i < nums.length; i++) { - rightSum = sum - leftSum - nums[i]; - if (leftSum == rightSum) { - return i; - } - leftSum += nums[i]; - } - return -1; - } - - public static void main(String[] args) { - int[] a = {1, 7, 3, 6, 5, 6}; - int[] b = {1, 2, 3}; - int[] c = {-1, -1, 0, 1, 0, -1}; - int[] d = {-1, -1, 0, 1, 1, 0}; - PivotIndex pivotIndex = new PivotIndex(); - System.out.println(Arrays.toString(a) + "的中心索引是:" + pivotIndex.pivotIndex(a)); - System.out.println(Arrays.toString(b) + "的中心索引是:" + pivotIndex.pivotIndex(b)); - System.out.println(Arrays.toString(c) + "的中心索引是:" + pivotIndex.pivotIndex(c)); - System.out.println(Arrays.toString(d) + "的中心索引是:" + pivotIndex.pivotIndex(d)); - } -} diff --git a/src/main/java/ds/ProductExceptSelf.java b/src/main/java/ds/ProductExceptSelf.java deleted file mode 100644 index 5796f68..0000000 --- a/src/main/java/ds/ProductExceptSelf.java +++ /dev/null @@ -1,45 +0,0 @@ -package ds; - -import java.util.Arrays; - -/** - * 除自身以外数组的乘积 - *
- *
- * 给定长度为 n 的整数数组 nums,其中 n> 1,返回输出数组 output ,其中 output[i] 等于 nums 中除 nums[i] 之外其余各元素的乘积。 - *
- * 示例: - *
- * 输入: [1,2,3,4] - * 输出: [24,12,8,6] - * 说明: 请不要使用除法,且在 O(n) 时间复杂度内完成此题。 - *
- * 进阶: - * 你可以在常数空间复杂度内完成这个题目吗?( 出于对空间复杂度分析的目的,输出数组不被视为额外空间。) - * - * @author yangyi 2019年03月02日22:54:06 - */ -public class ProductExceptSelf { - - public int[] productExceptSelf(int[] nums) { - int left = 1; - int right = 1; - int len = nums.length; - int[] output = new int[len]; - for (int i = 0; i < len; i++) { - output[i] = left; - left *= nums[i]; - } - for (int j = len - 1; j>= 0; j--) { - output[j] *= right; - right *= nums[j]; - } - return output; - } - - public static void main(String[] args) { - int[] a = {1,2,3,4}; - ProductExceptSelf productExceptSelf = new ProductExceptSelf(); - System.out.println(Arrays.toString(productExceptSelf.productExceptSelf(a))); - } -} diff --git a/src/main/java/ds/RepeatInArray.java b/src/main/java/ds/RepeatInArray.java deleted file mode 100644 index d7478d9..0000000 --- a/src/main/java/ds/RepeatInArray.java +++ /dev/null @@ -1,77 +0,0 @@ -package ds; - -import java.util.ArrayList; -import java.util.List; - -/** - * 给定一个整数数组 a,其中1 ≤ a[i] ≤ n (n为数组长度), 其中有些元素出现两次而其他元素出现一次。 - *
- * 找到所有出现两次的元素。 - *
- * 你可以不用到任何额外空间并在O(n)时间复杂度内解决这个问题吗? - *
- *
- *
- * 思路:
- * 1. 不能用额外空间,开辟散列表装数据的方法没戏。
- * 2. 时间复杂度为O(n),排序没戏。
- *
- * @author yangyi 2019年02月08日10:13:40
- */
-public class RepeatInArray {
-
-// public List
- * 虽然效率爆表,但是也有弊端,因为改变了数组的顺序,如果题目不要求效率转而要求数组的结构不变,则此方法不适用。
- */
- public List
- *
- *
- * 给定一个数组,将数组中的元素向右移动 k 个位置,其中 k 是非负数。
- *
- * 示例 1:
- *
- * 输入: [1,2,3,4,5,6,7] 和 k = 3
- * 输出: [5,6,7,1,2,3,4]
- * 解释:
- * 向右旋转 1 步: [7,1,2,3,4,5,6]
- * 向右旋转 2 步: [6,7,1,2,3,4,5]
- * 向右旋转 3 步: [5,6,7,1,2,3,4]
- * 示例 2:
- *
- * 输入: [-1,-100,3,99] 和 k = 2
- * 输出: [3,99,-1,-100]
- * 解释:
- * 向右旋转 1 步: [99,-1,-100,3]
- * 向右旋转 2 步: [3,99,-1,-100]
- * 说明:
- *
- * 尽可能想出更多的解决方案,至少有三种不同的方法可以解决这个问题。
- * 要求使用空间复杂度为 O(1) 的原地算法。
- *
- *
- *
- * 总共分为3步:
- *
- * 1. 数组整个旋转一遍
- * 2. 以k为界,k之前的元素旋转
- * 3. 以k为界,k后面的元素旋转
- *
- * @author yangyi 2019年02月12日11:56:52
- */
-public class RotateArray {
-
- private static int[] a = {1, 2, 3, 4, 5, 6, 7};
- private static int[] b = {-1, -100, 3, 99};
-
- public void rotate(int[] nums, int k) {
- int len = nums.length;
- k %= len;
- reserve(nums, 0, len - 1);
- reserve(nums, 0, k - 1);
- reserve(nums, k, len - 1);
- }
-
- public void reserve(int[] nums, int start, int end) {
- while (start < end) { - int temp = nums[start]; - nums[start++] = nums[end]; - nums[end--] = temp; - } - } - - public static void main(String[] args) { - RotateArray rotateArray = new RotateArray(); - rotateArray.rotate(a,3); - rotateArray.rotate(b,2); - System.out.println(Arrays.toString(a)); - System.out.println(Arrays.toString(b)); - } -} diff --git a/src/main/java/ds/SortedSquares.java b/src/main/java/ds/SortedSquares.java deleted file mode 100644 index 19d9203..0000000 --- a/src/main/java/ds/SortedSquares.java +++ /dev/null @@ -1,51 +0,0 @@ -package ds; - -import java.util.Arrays; - -/** - * 有序数组的平方 - *
- *
- * 给定一个按非递减顺序排序的整数数组 A,返回每个数字的平方组成的新数组,要求也按非递减顺序排序。
- *
- *
- *
- * 示例 1:
- *
- * 输入:[-4,-1,0,3,10]
- * 输出:[0,1,9,16,100]
- * 示例 2:
- *
- * 输入:[-7,-3,2,3,11]
- * 输出:[4,9,9,49,121]
- *
- *
- * 提示:
- *
- * 1 <= A.length <= 10000 - * -10000 <= A[i] <= 10000 - * A 已按非递减顺序排序。 - * - * @author yangyi 2019年02月16日23:17:48 - */ -public class SortedSquares { - - public int[] sortedSquares(int[] A) { - if (A == null || A.length == 0) { - return new int[]{}; - } - for (int i = 0; i < A.length; i++) { - A[i] *= A[i]; - } - Arrays.sort(A); - return A; - } - - public static void main(String[] args) { - int[] a = {-4, -1, 0, 3, 10}; - int[] b = {-7, -3, 2, 3, 11}; - SortedSquares sortedSquares = new SortedSquares(); - System.out.println(Arrays.toString(sortedSquares.sortedSquares(a))); - System.out.println(Arrays.toString(sortedSquares.sortedSquares(b))); - } -} diff --git a/src/main/java/ds/array/leetcode189/Solution.java b/src/main/java/ds/array/leetcode189/Solution.java new file mode 100644 index 0000000..e8e6459 --- /dev/null +++ b/src/main/java/ds/array/leetcode189/Solution.java @@ -0,0 +1,43 @@ +package ds.array.leetcode189; + +import java.util.Arrays; + +/** + * 189. 轮转数组 + * https://leetcode.cn/problems/rotate-array/ + * 总共分为3步: + * 1. 数组整个旋转一遍 + * 2. 以k为界,k之前的元素旋转 + * 3. 以k为界,k后面的元素旋转 + * + * @author yangyi 2023年09月04日15:53:56 + */ +public class Solution { + + private static int[] a = {1, 2, 3, 4, 5, 6, 7}; + private static int[] b = {-1, -100, 3, 99}; + + public void rotate(int[] nums, int k) { + int len = nums.length; + k %= len; + reserve(nums, 0, len - 1); + reserve(nums, 0, k - 1); + reserve(nums, k, len - 1); + } + + public void reserve(int[] nums, int start, int end) { + while (start < end) { + int temp = nums[start]; + nums[start++] = nums[end]; + nums[end--] = temp; + } + } + + public static void main(String[] args) { + Solution rotateArray = new Solution(); + rotateArray.rotate(a, 3); + rotateArray.rotate(b, 2); + System.out.println(Arrays.toString(a)); + System.out.println(Arrays.toString(b)); + } +} diff --git a/src/main/java/ds/ContainsNearbyDuplicate.java b/src/main/java/ds/array/leetcode219/Solution.java similarity index 57% rename from src/main/java/ds/ContainsNearbyDuplicate.java rename to src/main/java/ds/array/leetcode219/Solution.java index 7c9d369..dcdacbe 100644 --- a/src/main/java/ds/ContainsNearbyDuplicate.java +++ b/src/main/java/ds/array/leetcode219/Solution.java @@ -1,24 +1,12 @@ -package ds; - -import java.util.Arrays; +package ds.array.leetcode219; /** - * 给定一个整数数组和一个整数 k,判断数组中是否存在两个不同的索引 i 和 j,使得 nums [i] = nums [j],并且 i 和 j 的差的绝对值最大为 k。 - *
- * 示例 1:
- *
- * 输入: nums = [1,2,3,1], k = 3
- * 输出: true
- * 示例 2:
- *
- * 输入: nums = [1,0,1,1], k = 1
- * 输出: true
- * 示例 3:
- *
- * 输入: nums = [1,2,3,1,2,3], k = 2
- * 输出: false
+ * 219. 存在重复元素 II
+ * https://leetcode.cn/problems/contains-duplicate-ii/description/
+ *
+ * @author yangyi 2023年09月04日18:03:26
*/
-public class ContainsNearbyDuplicate {
+public class Solution {
public boolean containsNearbyDuplicate(int[] nums, int k) {
for (int i = 0; i < nums.length; i++) { @@ -35,7 +23,7 @@ public static void main(String[] args) { int[] nums_1 = {1, 2, 3, 1}; int[] nums_2 = {1, 0, 1, 1}; int[] nums_3 = {1, 2, 3, 1, 2, 3}; - ContainsNearbyDuplicate containsNearbyDuplicate = new ContainsNearbyDuplicate(); + Solution containsNearbyDuplicate = new Solution(); System.out.println(containsNearbyDuplicate.containsNearbyDuplicate(nums_1, 3)); System.out.println(containsNearbyDuplicate.containsNearbyDuplicate(nums_2, 1)); System.out.println(containsNearbyDuplicate.containsNearbyDuplicate(nums_3, 2)); diff --git a/src/main/java/ds/array/leetcode238/Solution.java b/src/main/java/ds/array/leetcode238/Solution.java new file mode 100644 index 0000000..d63b682 --- /dev/null +++ b/src/main/java/ds/array/leetcode238/Solution.java @@ -0,0 +1,34 @@ +package ds.array.leetcode238; + +import java.util.Arrays; + +/** + * 238. 除自身以外数组的乘积 + * https://leetcode.cn/problems/product-of-array-except-self/ + * + * @author yangyi 2023年09月04日17:31:30 + */ +public class Solution { + + public int[] productExceptSelf(int[] nums) { + int left = 1; + int right = 1; + int len = nums.length; + int[] output = new int[len]; + for (int i = 0; i < len; i++) { + output[i] = left; + left *= nums[i]; + } + for (int j = len - 1; j>= 0; j--) {
+ output[j] *= right;
+ right *= nums[j];
+ }
+ return output;
+ }
+
+ public static void main(String[] args) {
+ int[] a = {1,2,3,4};
+ Solution productExceptSelf = new Solution();
+ System.out.println(Arrays.toString(productExceptSelf.productExceptSelf(a)));
+ }
+}
diff --git a/src/main/java/ds/IncreasingTriplet.java b/src/main/java/ds/array/leetcode334/Solution.java
similarity index 52%
rename from src/main/java/ds/IncreasingTriplet.java
rename to src/main/java/ds/array/leetcode334/Solution.java
index 30a2fb2..ad7c461 100644
--- a/src/main/java/ds/IncreasingTriplet.java
+++ b/src/main/java/ds/array/leetcode334/Solution.java
@@ -1,31 +1,12 @@
-package ds;
+package ds.array.leetcode334;
/**
- * 递增的三元子序列
- *
- *
- *
- *
- * 给定一个未排序的数组,判断这个数组中是否存在长度为 3 的递增子序列。
- *
- * 数学表达式如下:
- *
- * 如果存在这样的 i, j, k, 且满足 0 ≤ i < j < k ≤ n-1, - * 使得 arr[i] < arr[j] < arr[k] ,返回 true ; 否则返回 false 。 - * 说明: 要求算法的时间复杂度为 O(n),空间复杂度为 O(1) 。 - *
- * 示例 1:
- *
- * 输入: [1,2,3,4,5]
- * 输出: true
- * 示例 2:
- *
- * 输入: [5,4,3,2,1]
- * 输出: false
+ * 334. 递增的三元子序列
+ * https://leetcode.cn/problems/increasing-triplet-subsequence/
*
* @author yangyi 2019年03月02日22:45:15
*/
-public class IncreasingTriplet {
+public class Solution {
public boolean increasingTriplet(int[] nums) {
int min = Integer.MAX_VALUE;
@@ -45,7 +26,7 @@ public boolean increasingTriplet(int[] nums) {
public static void main(String[] args) {
int[] a = {1,2,3,4,5};
int[] b = {5,4,3,2,1};
- IncreasingTriplet increasingTriplet = new IncreasingTriplet();
+ Solution increasingTriplet = new Solution();
System.out.println(increasingTriplet.increasingTriplet(a));
System.out.println(increasingTriplet.increasingTriplet(b));
}
diff --git a/src/main/java/ds/Intersect.java b/src/main/java/ds/array/leetcode350/Solution.java
similarity index 55%
rename from src/main/java/ds/Intersect.java
rename to src/main/java/ds/array/leetcode350/Solution.java
index 561c98f..6d33a91 100644
--- a/src/main/java/ds/Intersect.java
+++ b/src/main/java/ds/array/leetcode350/Solution.java
@@ -1,39 +1,14 @@
-package ds;
+package ds.array.leetcode350;
import java.util.*;
/**
- * 两个数组的交集 II
- *
- *
- * 给定两个数组,编写一个函数来计算它们的交集。
- *
- * 示例 1:
- *
- * 输入: nums1 = [1,2,2,1], nums2 = [2,2]
- * 输出: [2,2]
- * 示例 2:
- *
- * 输入: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
- * 输出: [4,9]
- *
- *
- * -------------------------------------------------
- *
- *
- * 说明:
- *
- * 输出结果中每个元素出现的次数,应与元素在两个数组中出现的次数一致。
- * 我们可以不考虑输出结果的顺序。
- * 进阶:
- *
- * 如果给定的数组已经排好序呢?你将如何优化你的算法?
- * 如果 nums1 的大小比 nums2 小很多,哪种方法更优?
- * 如果 nums2 的元素存储在磁盘上,磁盘内存是有限的,并且你不能一次加载所有的元素到内存中,你该怎么办?
+ * 350. 两个数组的交集 II
+ * https://leetcode.cn/problems/intersection-of-two-arrays-ii/description/
*
- * @author yangyi 2019年03月02日22:00:51
+ * @author yangyi 2023年09月04日17:17:39
*/
-public class Intersect {
+public class Solution {
public int[] intersect(int[] nums1, int[] nums2) {
//key为数字,value为次数
@@ -60,7 +35,7 @@ public int[] intersect(int[] nums1, int[] nums2) {
}
public static void main(String[] args) {
- Intersect intersect = new Intersect();
+ Solution intersect = new Solution();
int[] a = {1, 2, 2, 1};
int[] b = {2, 2};
int[] a2 = {4, 9, 5};
diff --git a/src/main/java/ds/array/leetcode442/Solution.java b/src/main/java/ds/array/leetcode442/Solution.java
new file mode 100644
index 0000000..baf62db
--- /dev/null
+++ b/src/main/java/ds/array/leetcode442/Solution.java
@@ -0,0 +1,43 @@
+package ds.array.leetcode442;
+
+import java.util.ArrayList;
+import java.util.List;
+
+/**
+ * 442. 数组中重复的数据
+ * https://leetcode.cn/problems/find-all-duplicates-in-an-array/
+ * 思路:
+ * 1. 不能用额外空间,开辟散列表装数据的方法没戏。
+ * 2. 时间复杂度为O(n),排序没戏。
+ *
+ * @author yangyi 2023年09月04日11:30:43
+ */
+public class Solution {
+
+ /**
+ * 时间复杂度:O(n)
+ * 空间复杂度:O(1)
+ *
+ * 虽然效率爆表,但是也有弊端,因为改变了数组的顺序,如果题目不要求效率转而要求数组的结构不变,则此方法不适用。
+ */
+ public List
- * 方法:
- *
- * 1. 从二维数组右上角的元素开始find循环查询,并以右上角元素为基准点。
- * 2. 如果要查询的元素比基准点小,基准点向左移。
- * 3. 如果要查询的元素比基准点大,基准点向下移。
+ * 74. 搜索二维矩阵
+ * https://leetcode.cn/problems/search-a-2d-matrix/
*
- * @author yangyi 2019年02月08日09:20:55
+ * @author yangyi 2023年09月04日11:14:35
*/
-public class FindInDoubleArray {
+public class Solution {
/**
* @param matrix 待查找的目标二维数组
@@ -60,7 +55,7 @@ public static void main(String[] args) {
}
System.out.println();
}
- FindInDoubleArray findInDoubleArray = new FindInDoubleArray();
+ Solution findInDoubleArray = new Solution();
int target1 = 6;
int target2 = 8;
int target3 = 0;
diff --git a/src/main/java/ds/array/leetcode88/Solution.java b/src/main/java/ds/array/leetcode88/Solution.java
new file mode 100644
index 0000000..faf3f7d
--- /dev/null
+++ b/src/main/java/ds/array/leetcode88/Solution.java
@@ -0,0 +1,28 @@
+package ds.array.leetcode88;
+
+import java.util.Arrays;
+
+/**
+ * 88.合并两个有序数组
+ * https://leetcode.cn/problems/merge-sorted-array/
+ *
+ * @author yangyi 2023年09月04日15:27:42
+ */
+public class Solution {
+
+ public void merge(int[] nums1, int m, int[] nums2, int n) {
+ for (int i = 0; i < n; i++) { + nums1[m + i] = nums2[i]; + } + Arrays.sort(nums1); + } + + public static void main(String[] args) { + int[] nums1 = {1, 2, 3, 0, 0, 0}; + int[] nums2 = {2, 5, 6}; + new Solution().merge(nums1, 3, nums2, 3); + System.out.println(Arrays.toString(nums1)); + System.out.println(Arrays.toString(nums2)); + } + +} diff --git a/src/main/java/ds/array/leetcode977/Solution.java b/src/main/java/ds/array/leetcode977/Solution.java new file mode 100644 index 0000000..6670f15 --- /dev/null +++ b/src/main/java/ds/array/leetcode977/Solution.java @@ -0,0 +1,31 @@ +package ds.array.leetcode977; + +import java.util.Arrays; + +/** + * 977. 有序数组的平方 + * https://leetcode.cn/problems/squares-of-a-sorted-array/ + * + * @author yangyi 2023年09月04日16:37:38 + */ +public class Solution { + + public int[] sortedSquares(int[] nums) { + if (nums == null || nums.length == 0) { + return new int[]{}; + } + for (int i = 0; i < nums.length; i++) { + nums[i] *= nums[i]; + } + Arrays.sort(nums); + return nums; + } + + public static void main(String[] args) { + int[] a = {-4, -1, 0, 3, 10}; + int[] b = {-7, -3, 2, 3, 11}; + Solution sortedSquares = new Solution(); + System.out.println(Arrays.toString(sortedSquares.sortedSquares(a))); + System.out.println(Arrays.toString(sortedSquares.sortedSquares(b))); + } +} diff --git a/src/main/java/ds/dfs/leetcode129/Solution.java b/src/main/java/ds/dfs/leetcode129/Solution.java new file mode 100644 index 0000000..475dc2b --- /dev/null +++ b/src/main/java/ds/dfs/leetcode129/Solution.java @@ -0,0 +1,54 @@ +package ds.dfs.leetcode129; + +/** + * 129. 求根节点到叶节点数字之和 + * https://leetcode.cn/problems/sum-root-to-leaf-numbers/description/?envType=study-plan-v2&envId=top-interview-150 + * + * @author yangyi 2023年09月04日01:32:43 + */ +class Solution { + + public static class TreeNode { + int val; + TreeNode left; + TreeNode right; + + TreeNode() { + } + + TreeNode(int val) { + this.val = val; + } + + TreeNode(int val, TreeNode left, TreeNode right) { + this.val = val; + this.left = left; + this.right = right; + } + } + + public int sumNumbers(TreeNode root) { + return sumNumbersDfs(root, 0); + } + + public int sumNumbersDfs(TreeNode root, int preSum) { + if (root == null) { + return 0; + } + int sum = preSum * 10 + root.val; + if (root.left == null && root.right == null) { + return sum; + } else { + return sumNumbersDfs(root.left, sum) + sumNumbersDfs(root.right, sum); + } + } + + public static void main(String[] args) { + TreeNode one = new TreeNode(1); + TreeNode two = new TreeNode(2); + TreeNode three = new TreeNode(3); + one.left = two; + one.right = three; + System.out.println(new Solution().sumNumbers(one)); + } +} diff --git a/src/main/java/ds/dp/leetcode322/Solution.java b/src/main/java/ds/dp/leetcode322/Solution.java new file mode 100644 index 0000000..666e6ba --- /dev/null +++ b/src/main/java/ds/dp/leetcode322/Solution.java @@ -0,0 +1,35 @@ +package ds.dp.leetcode322; + +/** + * 322. 零钱兑换 + * https://leetcode.cn/problems/coin-change/description/ + * + * @author yangyi 2023年09月13日00:54:39 + */ +class Solution { + + public int coinChange(int[] coins, int amount) { + if (amount == 0) { + return 0; + } + int[] dp = new int[amount + 1]; + for (int i = 1; i <= amount; i++) { + dp[i] = 999999; + for (int coin : coins) { + if (i - coin < 0) { + continue; + } + dp[i] = Math.min(dp[i], dp[i - coin] + 1); + } + } + return dp[amount] == 999999 ? -1 : dp[amount]; + } + + public static void main(String[] args) { + System.out.println(new Solution().coinChange(new int[]{1, 2, 5}, 11)); + System.out.println(new Solution().coinChange(new int[]{2}, 3)); + System.out.println(new Solution().coinChange(new int[]{1}, 0)); + } + + +} diff --git a/src/main/java/ds/dp/leetcode64/Solution.java b/src/main/java/ds/dp/leetcode64/Solution.java new file mode 100644 index 0000000..715e987 --- /dev/null +++ b/src/main/java/ds/dp/leetcode64/Solution.java @@ -0,0 +1,33 @@ +package ds.dp.leetcode64; + +/** + * 64. 最小路径和 + * https://leetcode.cn/problems/minimum-path-sum/description/?envType=study-plan-v2&envId=xiaohongshu-2023-fall-sprint + * + * @author yangyi 2023年09月03日21:36:30 + */ +class Solution { + + public int minPathSum(int[][] grid) { + int[][] dp = new int[grid.length][grid[0].length]; + for (int i = grid.length - 1; i>= 0; i--) {
+ for (int j = grid[0].length - 1; j>= 0; j--) {
+ if (i == grid.length - 1 && j != grid[0].length - 1) {
+ dp[i][j] = grid[i][j] + dp[i][j + 1];
+ } else if (i != grid.length - 1 && j == grid[0].length - 1) {
+ dp[i][j] = grid[i][j] + dp[i + 1][j];
+ } else if (i != grid.length - 1 && j != grid[0].length - 1) {
+ dp[i][j] = grid[i][j] + Math.min(dp[i][j + 1], dp[i + 1][j]);
+ } else {
+ dp[i][j] = grid[i][j];
+ }
+ }
+ }
+ return dp[0][0];
+ }
+
+ public static void main(String[] args) {
+ int[][] origin = {{1, 3, 1}, {1, 5, 1}, {4, 2, 1}};
+ System.out.println(new Solution().minPathSum(origin));
+ }
+}
diff --git a/src/main/java/ds/hashmap/leetcode409/Solution.java b/src/main/java/ds/hashmap/leetcode409/Solution.java
new file mode 100644
index 0000000..6e73e4d
--- /dev/null
+++ b/src/main/java/ds/hashmap/leetcode409/Solution.java
@@ -0,0 +1,31 @@
+package ds.hashmap.leetcode409;
+
+import java.util.HashSet;
+
+/**
+ * 409. 最长回文串
+ * https://leetcode.cn/problems/longest-palindrome/description/
+ *
+ * @author yangyi 2023年09月03日22:38:14
+ */
+class Solution {
+
+ public int longestPalindrome(String s) {
+ HashSet