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leecode更新

This commit is contained in:
markilue 2023-02-02 12:45:43 +08:00
parent c9c55164b7
commit 8805612c3a
10 changed files with 428 additions and 5 deletions
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2
Leecode/src/main/java/com/markilue/leecode/backtrace/Subsets.java → Leecode/src/main/java/com/markilue/leecode/backtrace/T08_Subsets.java
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@ -16,7 +16,7 @@ import java.util.List;
* 解集 不能 包含重复的子集。你可以按 任意顺序 返回解集。
* @Version: 1.0
*/
public class Subsets {
public class T08_Subsets {
@Test
public void test(){

2
Leecode/src/main/java/com/markilue/leecode/backtrace/SubsetsWithDup.java → Leecode/src/main/java/com/markilue/leecode/backtrace/T09_SubsetsWithDup.java
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@ -14,7 +14,7 @@ import java.util.*;
* 解集 不能 包含重复的子集。返回的解集中，子集可以按 任意顺序 排列。
* @Version: 1.0
*/
public class SubsetsWithDup {
public class T09_SubsetsWithDup {
@Test
public void test() {

2
Leecode/src/main/java/com/markilue/leecode/backtrace/FindSubsequences.java → Leecode/src/main/java/com/markilue/leecode/backtrace/T10_FindSubsequences.java
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@ -17,7 +17,7 @@ import java.util.Set;
* 数组中可能含有重复元素，如出现两个整数相等，也可以视作递增序列的一种特殊情况。
* @Version: 1.0
*/
public class FindSubsequences {
public class T10_FindSubsequences {
@Test

2
Leecode/src/main/java/com/markilue/leecode/backtrace/Permute.java → Leecode/src/main/java/com/markilue/leecode/backtrace/T11_Permute.java
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@ -15,7 +15,7 @@ import java.util.List;
* 给定一个不含重复数字的数组 nums ，返回其 所有可能的全排列 。你可以 按任意顺序 返回答案。
* @Version: 1.0
*/
public class Permute {
public class T11_Permute {
@Test
public void test(){

2
Leecode/src/main/java/com/markilue/leecode/backtrace/PermuteUnique.java → Leecode/src/main/java/com/markilue/leecode/backtrace/T12_PermuteUnique.java
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@ -16,7 +16,7 @@ import java.util.List;
* 给定一个可包含重复数字的序列 nums ，按任意顺序 返回所有不重复的全排列。
* @Version: 1.0
*/
public class PermuteUnique {
public class T12_PermuteUnique {
@Test
public void test() {

52
Leecode/src/main/java/com/markilue/leecode/backtrace/second/T08_Subsets.java Normal file
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@ -0,0 +1,52 @@
package com.markilue.leecode.backtrace.second;
import org.junit.Test;
import java.util.ArrayList;
import java.util.List;
/**
*@BelongsProject: Leecode
*@BelongsPackage: com.markilue.leecode.backtrace.second
*@Author: dingjiawen
*@CreateTime: 2023-02-02 10:07
*@Description:
* TODO 二刷力扣78题 子集:
* 给你一个整数数组 nums ，数组中的元素 互不相同 。返回该数组所有可能的子集（幂集）。
* 解集 不能 包含重复的子集。你可以按 任意顺序 返回解集。
*@Version: 1.0
*/
public class T08_Subsets {
@Test
public void test() {
int[] nums={1,2,3};
System.out.println(subsets(nums));
}
List<List<Integer>> result = new ArrayList<>();
List<Integer> cur = new ArrayList<>();
public List<List<Integer>> subsets(int[] nums) {
backtracking(nums, 0);
return result;
}
public void backtracking(int[] nums, int level) {
if (level == nums.length) {
result.add(new ArrayList<>(cur));
return;
}
for (int i = 0; i <= 1; i++) {
if (i == 0) {
backtracking(nums, level + 1);
} else {
cur.add(nums[level]);
backtracking(nums, level+1);
cur.remove(cur.size() - 1);
}
}
}
}

100
Leecode/src/main/java/com/markilue/leecode/backtrace/second/T09_SubsetsWithDup.java Normal file
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@ -0,0 +1,100 @@
package com.markilue.leecode.backtrace.second;
import org.junit.Test;
import java.util.*;
/**
*@BelongsProject: Leecode
*@BelongsPackage: com.markilue.leecode.backtrace.second
*@Author: dingjiawen
*@CreateTime: 2023-02-02 10:32
*@Description:
* TODO 力扣90题 子集II:
* 给你一个整数数组 nums ，其中可能包含重复元素，请你返回该数组所有可能的子集（幂集）。
* 解集 不能 包含重复的子集。返回的解集中，子集可以按 任意顺序 排列。
*@Version: 1.0
*/
public class T09_SubsetsWithDup {
@Test
public void test() {
int[] nums = {1, 2, 2};
System.out.println(subsetsWithDup1(nums));
}
/**
* 两种思路:1.树层去重2.map记录元素个数，横向遍历使用
* @param nums
* @return
*/
Map<Integer, Integer> map = new HashMap<Integer, Integer>();//<num,count>
List<Integer> key = new ArrayList<>();
List<Integer> cur = new ArrayList<>();
List<List<Integer>> result = new ArrayList<>();
public List<List<Integer>> subsetsWithDup(int[] nums) {
for (int num : nums) {
Integer count = map.getOrDefault(num, 0);
if (count == 0) {
key.add(num);
}
map.put(num, count + 1);
}
backtracking(0);
return result;
}
/**
* map记录次数法
* @param level
*/
public void backtracking(int level) {
if (level == key.size()) {
result.add(new ArrayList<>(cur));
return;
}
Integer keyNow = key.get(level);
Integer count = map.get(keyNow);
for (int i = 0; i <= count; i++) {
if (i != 0) {
cur.add(keyNow);
}
backtracking(level + 1);
}
for (int i = 0; i < count; i++) {
cur.remove(cur.size() - 1);
}
}
public List<List<Integer>> subsetsWithDup1(int[] nums) {
Arrays.sort(nums);
backtracking1(nums,0);
return result;
}
/**
* 树层去重法
* @param nums
*/
public void backtracking1(int[] nums, int startIndex) {
result.add(new ArrayList<>(cur));
if (startIndex > nums.length) {
return;
}
for (int i = startIndex; i < nums.length; i++) {
//树层去重
if (i != startIndex && nums[i - 1] == nums[i]) {
continue;
}
cur.add(nums[i]);
backtracking1(nums, i + 1);
cur.remove(cur.size() - 1);
}
}
}

101
Leecode/src/main/java/com/markilue/leecode/backtrace/second/T10_FindSubsequences.java Normal file
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@ -0,0 +1,101 @@
package com.markilue.leecode.backtrace.second;
import org.junit.Test;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
/**
*@BelongsProject: Leecode
*@BelongsPackage: com.markilue.leecode.backtrace.second
*@Author: dingjiawen
*@CreateTime: 2023-02-02 11:28
*@Description:
* TODO 二刷力扣491 递增子序列:
* 给你一个整数数组 nums ，找出并返回所有该数组中不同的递增子序列，递增子序列中 至少有两个元素 。你可以按 任意顺序 返回答案。
* 数组中可能含有重复元素，如出现两个整数相等，也可以视作递增序列的一种特殊情况。
*@Version: 1.0
*/
public class T10_FindSubsequences {
@Test
public void test() {
int[] nums = {4, 6, 7, 7};
System.out.println(findSubsequences(nums));
}
List<Integer> cur = new ArrayList<>();
List<List<Integer>> result = new ArrayList<>();
/**
* 与上一题子集II类似，都是需要树层去重即可
* @param nums
* @return
*/
public List<List<Integer>> findSubsequences(int[] nums) {
backtracking(nums, 0);
return result;
}
public void backtracking(int[] nums, int startIndex) {
if (cur.size() >= 2) {
result.add(new ArrayList<>(cur));
}
if (startIndex == nums.length) return;
HashSet<Integer> set = new HashSet<>();
for (int i = startIndex; i < nums.length; i++) {
//同一树层去除,由于相同数不一定连续,所以使用set来进行树层去重
//这里还可以使用数组来去重，当数组的指定数位置的值为1时则证明使用过
if (set.contains(nums[i])) {
continue;
}
set.add(nums[i]);
if (cur.isEmpty() || cur.get(cur.size() - 1) <= nums[i]) {
//是递增的
cur.add(nums[i]);
backtracking(nums, i + 1);
cur.remove(cur.size() - 1);
}
}
}
/**
* 官方最快 2ms,可能快在用lastNum来记录cur(last)
* @param nums
* @return
*/
public List<List<Integer>> findSubsequences1(int[] nums) {
List<List<Integer>> list = new ArrayList<>();
if (nums == null || nums.length == 0) return list;
List<Integer> result = new ArrayList<>();
dfs(0, nums, result, list, -101);
return list;
}
private void dfs(int idx, int[] nums, List<Integer> result, List<List<Integer>> list, int lastNum) {
if (result.size() > 1) {
list.add(new ArrayList<>(result));
}
for (int i = idx; i < nums.length; i++) {
if (isRepeat(nums, idx, i)) continue;
if (nums[i] < lastNum) continue;
result.add(nums[i]);
dfs(i + 1, nums, result, list, nums[i]);
result.remove(result.size() - 1);
}
}
private boolean isRepeat(int[] nums, int idx, int i) {
for (int j = idx; j < i; j++) {
if (nums[j] == nums[i]) return true;
}
return false;
}
}

83
Leecode/src/main/java/com/markilue/leecode/backtrace/second/T11_Permute.java Normal file
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@ -0,0 +1,83 @@
package com.markilue.leecode.backtrace.second;
import com.sun.org.apache.bcel.internal.generic.LUSHR;
import java.util.ArrayList;
import java.util.List;
/**
*@BelongsProject: Leecode
*@BelongsPackage: com.markilue.leecode.backtrace.second
*@Author: dingjiawen
*@CreateTime: 2023-02-02 11:55
*@Description:
* TODO 力扣46题 全排列:
* 给定一个不含重复数字的数组 nums ，返回其 所有可能的全排列 。你可以 按任意顺序 返回答案。
*@Version: 1.0
*/
public class T11_Permute {
List<Integer> cur = new ArrayList<>();
List<List<Integer>> result = new ArrayList<>();
boolean[] used;
/**
* 全排列的题，实际上就是顺序也有关系，所以直接不用传startIndex,每次都从0开始就好
* 把used放外面速度击败82.16%，内存击败28.47% 1ms
* @param nums
* @return
*/
public List<List<Integer>> permute(int[] nums) {
used = new boolean[nums.length];
backtracking(nums);
return result;
}
public void backtracking(int[] nums) {
if(cur.size()==nums.length){
result.add(new ArrayList<>(cur));
return;
}
for (int i = 0; i < nums.length; i++) {
if(!used[i]){
used[i]=true;
cur.add(nums[i]);
backtracking(nums);
cur.remove(cur.size()-1);
used[i]=false;
}
}
}
/**
* 把used放里面传参
* 速度击败100%，内存击败83.99% 0ms
* @param nums
* @return
*/
public List<List<Integer>> permute1(int[] nums) {
backtracking(nums,new boolean[nums.length]);
return result;
}
public void backtracking(int[] nums, boolean[] used) {
if(cur.size()==nums.length){
result.add(new ArrayList<>(cur));
return;
}
for (int i = 0; i < nums.length; i++) {
if(!used[i]){
used[i]=true;
cur.add(nums[i]);
backtracking(nums,used);
cur.remove(cur.size()-1);
used[i]=false;
}
}
}
}

87
Leecode/src/main/java/com/markilue/leecode/backtrace/second/T12_PermuteUnique.java Normal file
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@ -0,0 +1,87 @@
package com.markilue.leecode.backtrace.second;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
/**
*@BelongsProject: Leecode
*@BelongsPackage: com.markilue.leecode.backtrace.second
*@Author: dingjiawen
*@CreateTime: 2023-02-02 12:08
*@Description:
* TODO 力扣47题 全排列II:
* 给定一个可包含重复数字的序列 nums ，按任意顺序 返回所有不重复的全排列。
*@Version: 1.0
*/
public class T12_PermuteUnique {
List<Integer> cur = new ArrayList<>();
List<List<Integer>> result = new ArrayList<>();
/**
* 要做的事似乎还是树层去重
* @param nums
* @return
*/
public List<List<Integer>> permuteUnique(int[] nums) {
backtracking(nums, new boolean[nums.length]);
return result;
}
public void backtracking(int[] nums, boolean[] used) {
if (nums.length == cur.size()) {
result.add(new ArrayList<>(cur));
return;
}
boolean[] curUsed = new boolean[21];//本树层是否用过
for (int i = 0; i < nums.length; i++) {
if (curUsed[nums[i] + 10]) continue;
if (!used[i]) {
curUsed[nums[i] + 10] = true;
used[i] = true;
cur.add(nums[i]);
backtracking(nums, used);
cur.remove(cur.size() - 1);
used[i] = false;
}
}
}
/**
* 可以进一步优化为一个used数组
* @param nums
* @param
*/
public List<List<Integer>> permuteUnique1(int[] nums) {
Arrays.sort(nums);
backtracking(nums, new boolean[nums.length]);
return result;
}
public void backtracking1(int[] nums, boolean[] used) {
if (nums.length == cur.size()) {
result.add(new ArrayList<>(cur));
return;
}
for (int i = 0; i < nums.length; i++) {
// used[i - 1] == true，说明同一树枝nums[i - 1]使用过
// used[i - 1] == false，说明同一树层nums[i - 1]使用过
// 如果同一树层nums[i - 1]使用过则直接跳过
if (i > 0 && nums[i] == nums[i - 1] && used[i - 1] == false) {
continue;
}
if (!used[i]) {
used[i] = true;
cur.add(nums[i]);
backtracking1(nums, used);
cur.remove(cur.size() - 1);
used[i] = false;
}
}
}
}
}