This commit is contained in:
whai 2024-03-16 19:02:55 +08:00
parent b633f27b27
commit 99bda3a447
13 changed files with 818 additions and 1 deletions

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@ -0,0 +1,67 @@
package cn.whaifree.leetCode.BackTracking;
import org.junit.Test;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
/**
* @version 1.0
* @Author whai文海
* @Date 2024/3/15 14:51
* @注释
*/
public class LeetCode698 {
@Test
public void test() {
int[] nums = {4, 3, 2, 3, 5, 2, 1};
int k = 4;
System.out.println(new Solution().canPartitionKSubsets(nums, k));
}
class Solution {
public int[] numUsed;
public boolean canPartitionKSubsets(int[] nums, int k) {
numUsed = new int[nums.length];
Arrays.sort(nums);
int sum = Arrays.stream(nums).sum();
if (sum % k != 0 || nums[nums.length - 1] > sum / k)
return false;
return divideGroups(nums, nums.length - 1, sum / k, 0, k);
}
/**
*
* @param nums
* @param start
* @param target
* @param current
* @param k
* @return
*/
public boolean divideGroups(int[] nums, int start, int target, int current, int k) {
if (k == 1)
return true; // 分组操作执行k-1次之后最后剩余的元素就是最后一组了不需要再匹配
if (current == target)
return divideGroups(nums, nums.length - 1, target, 0, k - 1); // 分组操作执行k-1次后最后剩余的元素就是最后一组了不需要再匹配
for (int i = start; i >= 0; i--) {
if (numUsed[i] == 1 || current + nums[i] > target) continue; // 被使用的元素不能再次使用总和大于目标值也不能使用
numUsed[i] = 1; // 标记占用
if (divideGroups(nums, i - 1, target, current + nums[i], k)) return true;
numUsed[i] = 0; // 撤销标记
while (i > 0 && nums[i - 1] == nums[i]) i--; // 例如12333333...假如最右侧的3这个值没有匹配上那么它左侧的剩余五个3都不需要再匹配了
}
return false;
}
}
}

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@ -12,10 +12,29 @@ public class BeiBao {
@Test @Test
public void main() { public void main() {
//
// new Thread(() -> {
// try {
// Thread.sleep(1000);
// } catch (InterruptedException e) {
// throw new RuntimeException(e);
// }
// }).start();
//
// new Thread(() -> {
//
// }).start();
//
//
// String a = "a";
//
// String b = new String("a");
//
// System.out.println(a == b);
int[] weight = {1,3,4}; int[] weight = {1,3,4};
int[] value = {15,20,30}; int[] value = {15,20,30};
int bagSize = 4; int bagSize = 4;
new Solution().packageProblem(weight, value, bagSize); new Solution2().packageProblem(weight, value, bagSize);
} }
class Solution{ class Solution{
@ -81,4 +100,106 @@ public class BeiBao {
} }
} }
class Solution1{
/**
* dp[i][j] i表示商品j表示空间 dp[i][j]表示 容量为j为包裹从0-i个商品中取得最大值
*
* 初始化
* 0 1 2 3 4 5 包容量
* 物品0 0 0 2 2 2 2
* 物品1 0
* 物品2 0
*
* dp[i][j] 表示从下标为[0-i]的物品里任意取放进容量为j的背包价值总和最大是多少
*
* @param weights 物品的重量
* @param values 物品的价值
* @param carryNumber 可以携带的数量
* @return
*/
public int packageProblem(int[] weights, int[] values, int carryNumber) {
int length = weights.length;
int[][] dp = new int[length][carryNumber + 1];
for (int i = 0; i < length; i++) {
dp[i][0] = 0;
}
for (int i = weights[0]; i <= carryNumber; i++) {
dp[0][i] = values[0];
}
// 如果i 不放东西 dp[i][j] = dp[i-1][j]
// 如果i 放东西 dp[i][j] = max(dp[i-1][j],dp[i-1][j-weight[i]]-value[i])
// i为商品j为容量
for (int i = 1; i < length; i++) {
for (int j = 1; j <= carryNumber; j++) {
if (weights[i] > j) {
// 重量太大放不进去
dp[i][j] = dp[i - 1][j];
}else {
dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - weights[i]] + values[i]);
}
}
}
return dp[length - 1][carryNumber];
}
}
class Solution2{
/**
*
* dp[i][j] 表示从下标为[0-i]的物品里任意取放进容量为j的背包价值总和最大是多少
* dp[j] 表示容量为j的背包的最大价值总和
*
* 初始化
* 0 1 2 3 4 5 包容量
* 物品0 0 0 2 2 2 2
* 物品1 0
* 物品2 0
*
* 不放这个商品 那价值不变还是dp[j]
* dp[j] = max(dp[j] , dp[j-weight[i]]+value[i]])
*
* @param weights 物品的重量
* @param values 物品的价值
* @param carryNumber 可以携带的数量
* @return
*/
public int packageProblem(int[] weights, int[] values, int carryNumber) {
int length = weights.length;
int[] dp = new int[carryNumber + 1];
// 不放这个商品 那价值不变还是dp[j]
// dp[j] = max(dp[j] , dp[j-weight[i]]+value[i]])
// i为商品j为容量
for(int i = 0; i < length; i++) { // 遍历物品
for(int j = carryNumber ;j >= weights[i]; j--) { // 遍历背包容量
dp[j] = Math.max(dp[j], dp[j - weights[i]] + values[i]);
}
/**
* 倒序遍历是为了保证物品i只被放入一次但如果一旦正序遍历了那么物品0就会被重复加入多次
*
* 举一个例子物品0的重量weight[0] = 1价值value[0] = 15
* 如果正序遍历
* dp[1] = dp[1 - weight[0]] + value[0] = 15
* dp[2] = dp[2 - weight[0]] + value[0] = 30
*
* 此时dp[2]就已经是30了意味着物品0被放入了两次所以不能正序遍历
*/
}
return dp[carryNumber];
}
}
} }

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@ -0,0 +1,207 @@
package cn.whaifree.leetCode.Dynamic;
import org.junit.Test;
import java.util.Arrays;
/**
* @version 1.0
* @Author whai文海
* @Date 2024/3/14 12:24
* @注释
*/
public class LeetCode416 {
@Test
public void test() {
int[] nums = {1,2,3,5};
boolean canPartition = new Solution3().canPartition(nums);
System.out.println(canPartition);
}
// class Solution {
// /**
// * 分割为2部分
// * @param nums
// * @return
// */
// public boolean canPartition(int[] nums) {
// Arrays.sort(nums);
//
// int rightSum = 0;
// for (int num : nums) {
// rightSum += num;
// }
// int leftSum = 0;
// for (int i = 0; i < nums.length; i++) {
// if (leftSum == rightSum) {
// return true;
// } else if (leftSum > rightSum) {
// return false;
// }
// leftSum += nums[i];
// rightSum -= nums[i];
// }
// return false;
// }
// }
class Solution {
int sumHalf = 0;
int nowSum = 0;
/**
* 回溯
*
* @param nums
* @return
*/
public boolean canPartition(int[] nums) {
for (int num : nums) {
sumHalf += num;
}
sumHalf /= 2;
return backTracking(nums, 0);
}
public boolean backTracking(int[] nums, int start) {
if (nowSum == sumHalf) {
return true;
}else if (nowSum>sumHalf){
return false;
}
boolean b = false;
for (int i = start; i < nums.length; i++) {
nowSum += nums[i];
b = backTracking(nums, i + 1);
// 不行有可能某些元素没在集合内
nowSum -= nums[i];
if (b==true) return true;
}
return b;
}
}
class Solution2 {
/**
* 转换为背包问题
* nums 为商品的重量同时为商品的价值
* 背包容量为sum/2 且需要装满
*
* 0 1 2 3 4 5
* 0 0 n0 n0 n0 n0 n0
* 1 0
* 2 0
* 3 0
* 4 0
* 5 0
*
* 不放东西的最大价值 dp[i][j] = dp[i-1][j]
* 放东西的最大价值 max(dp[i-1][j] , dp[i-1][j-nums[i]]+values[i])
*
* @param nums
* @return
*/
public boolean canPartition(int[] nums) {
int sumHalf = 0;
for (int num : nums) {
sumHalf += num;
}
if (sumHalf % 2 != 0) {
return false;
}
sumHalf /= 2;
int length = nums.length;
int[][] dp = new int[length][sumHalf + 1];
// for (int i = 0; i < length; i++) {
// dp[i][0] = 0;
// }
for (int i = nums[0]; i <= sumHalf; i++) {
dp[0][i] = nums[0];
}
for (int i = 1; i < length; i++) {
for (int j = 1; j <= sumHalf; j++) {
if (nums[i] > j) {
// 放不下
dp[i][j] = dp[i - 1][j];
} else {
dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - nums[i]] + nums[i]);
}
if (dp[i][j] == sumHalf) {
return true;
}
}
}
return false;
}
}
class Solution3 {
/**
* 将nums的元素放入背包中
* 1. 背包容量 sum/2
* 2. 物品i重量nums[i]价值nums[i]
*
* 0 1 2 3 4 5
* 0
* 1
* 2
*
* dp[j]表示 j背包容量的最大价值
*
*
*
* // 递推公式
* 2. 放下 dp[j] = maxdp[j], dp[j-weight[i]]+nums[i]
*
* 从后往前循环每次取得状态不会和之前取得状态重合这样每种物品就只取一次了
*
* @param nums
* @return
*/
public boolean canPartition(int[] nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
if (sum % 2 != 0) {
return false;
}
// 包裹最大容量
int packageCapacity = sum / 2;
int[] dp = new int[packageCapacity + 1];
for (int i = 0; i < nums.length; i++) {
// 包的容量必须大于物品i重量才能放进去
/**
* 如果正序遍历
* dp[1] = dp[1 - weight[0]] + value[0] = 15
* 此时容量1的背包已经放入了
* dp[2] = dp[2 - weight[0]] + value[0] = 30
* 此时又放入了一次
*/
for (int j = packageCapacity; j >= nums[i]; j--) {
dp[j] = Math.max(dp[j], dp[j - nums[i]] + nums[i]);
//剪枝一下每一次完成內層的for-loop立即檢查是否dp[target] == target優化時間複雜度26ms -> 20ms
if(dp[packageCapacity] == packageCapacity)
return true;
}
}
// 包裹容量和目标值一样
return dp[packageCapacity] == packageCapacity;
}
}
}

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@ -166,4 +166,48 @@ public class Tes1 {
t2.start(); t2.start();
} }
} }
class Sun implements Inter,Outer {
public static void main(String[] args) {
new Sun().testDefault();
Inter.testStatic();
}
/**
* 必须重写
*/
@Override
public void testDefault() {
Outer.super.testDefault();
Inter.super.testDefault();
}
}
interface Inter{
default void testDefault(){
System.out.println("testDefault");
}
static void testStatic(){
System.out.println("testStatic");
}
}
interface Outer{
default void testDefault(){
System.out.println("testDefault outer");
}
static void testStatic(){
System.out.println("testStatic outer");
}
}

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@ -0,0 +1,42 @@
package cn.whaifree.redo.redo_24_3_16;
/**
* @version 1.0
* @Author whai文海
* @Date 2024/3/16 16:59
* @注释
*/
public class LeetCode343 {
class Solution {
/**
* dp[j] 表示拆分出j后的乘积最大值
*
* dp[j]=
* (j-i)*j 两个数相乘
* dp[j-i] * i
*
* dp[j] = dp[j-1] *
* @param n
* @return
*/
public int integerBreak(int n) {
int[] dp = new int[n + 1];
dp[0] = 1;
dp[1] = 1;
dp[2] = 1;
for (int i = 3; i <= n; i++) {
for (int j = 1; j < i - 1; j++) {
dp[i] = Math.max(
dp[i],
Math.max((i - j) * j, dp[i - j] * j)
);
}
}
return dp[n];
}
}
}

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@ -0,0 +1,59 @@
package cn.whaifree.redo.redo_24_3_16;
import org.junit.Test;
import java.util.Arrays;
import java.util.Comparator;
/**
* @version 1.0
* @Author whai文海
* @Date 2024/3/16 17:42
* @注释
*/
public class LeetCode435 {
@Test
public void test() {
// int[][] intervals = {{1,2},{2,3},{3,4},{1,3}};
// int result = new Solution().eraseOverlapIntervals(intervals);
// System.out.println(result);
//
//
// // [ [1,2], [1,2], [1,2] ]
// int[][] intervals1 = {{1,2},{1,2},{1,2}};
// int result1 = new Solution().eraseOverlapIntervals(intervals1);
// System.out.println(result1);
// [[1,100],[11,22],[1,11],[2,12]]
int[][] intervals2 = {{1,100},{11,22},{1,11},{2,12}};
int result2 = new Solution().eraseOverlapIntervals(intervals2);
System.out.println(result2);
}
class Solution {
public int eraseOverlapIntervals(int[][] intervals) {
// 最多能参加几次活动
Arrays.sort(intervals, new Comparator<int[]>() {
@Override
public int compare(int[] o1, int[] o2) {
return o1[0] - o2[0];
}
});
int participation = 1;
for (int i = 1; i < intervals.length; i++) {
if (intervals[i][0] < intervals[i - 1][1]) {
// 后一个活动的开始时间早于头一个活动开始时间这个活动不能参加
intervals[i][1] = Math.min(intervals[i - 1][1], intervals[i][1]);
}else {
participation++;
}
}
return intervals.length - participation;
}
}
}

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@ -0,0 +1,45 @@
package cn.whaifree.redo.redo_24_3_16;
import org.junit.Test;
/**
* @version 1.0
* @Author whai文海
* @Date 2024/3/16 18:27
* @注释
*/
public class LeetCode45 {
@Test
public void test() {
int[] nums = new int[]{2, 3, 1, 1, 4};
int jump = new Solution().jump(nums);
System.out.println(jump);
}
class Solution {
public int jump(int[] nums) {
if (nums.length == 1) {
return 0;
}
int jump = 0;
int maxCover = 0;
int curCover = 0; // 用于标记是否达到一次跳的终点
for (int i = 0; i < nums.length; i++) {
maxCover = Math.max(maxCover, nums[i] + i);
if (i == curCover) {
jump++;
curCover = maxCover;
}
if (curCover >= nums.length - 1) {
return jump;
}
}
return jump;
}
}
}

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@ -0,0 +1,46 @@
package cn.whaifree.redo.redo_24_3_16;
import org.junit.Test;
/**
* @version 1.0
* @Author whai文海
* @Date 2024/3/16 18:04
* @注释
*/
public class LeetCode738 {
@Test
public void test() {
Solution solution = new Solution();
int n = 668799 ;
int n1 = 668799;
System.out.println(solution.monotoneIncreasingDigits(n));
}
class Solution {
public int monotoneIncreasingDigits(int n) {
// 668841 668799
// 遍历的过程-1 再找到最后一个递减的然后 后面全部替换为9
char[] chars = String.valueOf(n).toCharArray();
int index = Integer.MAX_VALUE;
for (int i = chars.length - 1; i > 0; i--) {
if (chars[i] < chars[i - 1]) {
chars[i-1]--;
index = i;
}
}
for (int i = index; i < chars.length ; i++) {
chars[i] = '9';
}
return Integer.parseInt(new String(chars));
}
}
}

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@ -0,0 +1,51 @@
package cn.whaifree.redo.redo_24_3_16;
import org.junit.Test;
import java.util.ArrayList;
import java.util.List;
/**
* @version 1.0
* @Author whai文海
* @Date 2024/3/16 17:53
* @注释
*/
public class LeetCode763 {
@Test
public void test() {
String s = "ababcbacadefegdehijhklij";
Solution solution = new Solution();
List<Integer> list = solution.partitionLabels(s);
System.out.println(list);
}
class Solution {
public List<Integer> partitionLabels(String s) {
// 统计每个字符最后出现的位置
int[] map = new int[26];
char[] chars = s.toCharArray();
for (int i = 0; i < chars.length; i++) {
map[chars[i] - 'a'] = i;
}
List<Integer> list = new ArrayList<>();
int left = 0;
int rightMax = 0;
for (int i = 0; i < chars.length; i++) {
rightMax = Math.max(map[chars[i] - 'a'], rightMax);
if (i == map[chars[i] - 'a'] && rightMax == i) {
list.add(rightMax - left + 1);
left = rightMax + 1;
}
}
return list;
}
}
}

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@ -0,0 +1,41 @@
package cn.whaifree.redo.redo_24_3_16;
import org.junit.Test;
/**
* @version 1.0
* @Author whai文海
* @Date 2024/3/16 16:40
* @注释
*/
public class LeetCode96 {
@Test
public void test() {
int n = 4;
int result = new Solution().numTrees(n);
System.out.println(result);
}
class Solution {
public int numTrees(int n) {
if (n == 1) {
return 1;
}
//dp[i]表示i个节点的可能的树的数量
int[] dp = new int[n + 1];
dp[0] = 1;
dp[1] = 1;
dp[2] = 2;
for (int i = 3; i <= n; i++) {
for (int j = 1; j <= i; j++) {
dp[i] += dp[i - j] * dp[j - 1];
}
}
return dp[n];
}
}
}

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@ -0,0 +1,64 @@
package cn.whaifree.redo.redo_24_3_16;
import cn.whaifree.leetCode.model.TreeNode;
import org.junit.Test;
/**
* @version 1.0
* @Author whai文海
* @Date 2024/3/16 18:18
* @注释
*/
public class LeetCode968 {
@Test
public void test() {
TreeNode treeNode = TreeNode.constructTreeByArray(0,0,null,0,0);
System.out.println(new Solution().minCameraCover(treeNode));
}
class Solution {
int res = 0;
/**
* 0 被覆盖
* 1 有监控
* 2 没覆盖
*
* @param root
* @return
*/
public int minCameraCover(TreeNode root) {
if (circle(root) == 2) {
res++;
}
return res;
}
public int circle(TreeNode root) {
if (root == null) {
return 0;
}
int left = circle(root.left);
int right = circle(root.right);
// 左边或者右边有未覆盖的就在这里安排一个监控
if (left == 2 || right == 2) {
res++;
return 1;
}
// 左边和右边有监控这里就被覆盖了
if (left == 1 || right == 1) {
return 0;
}
// 左右都覆盖这里就没覆盖
if (left == 0 && right == 0) {
return 2;
}
return -1;
}
}
}

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@ -0,0 +1,13 @@
package cn.whaifree.test;
/**
* @version 1.0
* @Author whai文海
* @Date 2024/3/14 14:58
* @注释
*/
@FunctionalInterface
public interface Custom {
void f();
}

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@ -0,0 +1,17 @@
package cn.whaifree.test;
/**
* @version 1.0
* @Author whai文海
* @Date 2024/3/14 14:58
* @注释
*/
public class LambdaClass {
public static void main(String[] args) {
new LambdaClass().lambdaInterfaceDemo(()-> System.out.println("自定义函数式接口"));
}
//函数式接口参数
void lambdaInterfaceDemo(Custom i){
i.f();
}
}