（算法题）Day-20230317
分割回文串
难点：
- 切割问题 可以抽象为组合问题
- 如何模拟那些切割线
- 切割问题中递归如何终止
- 在递归循环中如何截取子串
- 如何判断回文
package day.d0317;
import java.util.ArrayList;
import java.util.Deque;
import java.util.LinkedList;
import java.util.List;
/**
* @author hdbing
* @create 2023-03-17 13:58
*/
public class T01 {
public static void main(String[] args) {
Solution solution = new Solution();
List<List<String>> res = solution.partition("aab");
System.out.println(res);
}
}
class Solution {
List<List<String>> lists = new ArrayList<>();
Deque<String> deque = new LinkedList<>();
public List<List<String>> partition(String s) {
backTracking(s, 0);
return lists;
}
private void backTracking(String s, int startIndex) {
//如果起始位置大于s的大小，说明找到了一组分割方案
if (startIndex >= s.length()) {
lists.add(new ArrayList(deque));
return;
}
for (int i = startIndex; i < s.length(); i++) {
//如果是回文子串，则记录
if (isPalindrome(s, startIndex, i)) {
String str = s.substring(startIndex, i + 1);
deque.addLast(str);
} else {
continue;
}
//起始位置后移，保证不重复
backTracking(s, i + 1);
deque.removeLast();
}
}
//判断是否是回文串
private boolean isPalindrome(String s, int startIndex, int end) {
for (int i = startIndex, j = end; i < j; i++, j--) {
if (s.charAt(i) != s.charAt(j)) {
return false;
}
}
return true;
}
}
复原 IP 地址
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Deque;
import java.util.List;
import java.util.Stack;
public class Solution {
public List<String> restoreIpAddresses(String s) {
int len = s.length();
List<String> res = new ArrayList<>();
// 如果长度不够，不搜索
if (len < 4 || len > 12) {
return res;
}
Deque<String> path = new ArrayDeque<>(4);
int splitTimes = 0;
dfs(s, len, splitTimes, 0, path, res);
return res;
}
/**
* 判断 s 的子区间 [left, right] 是否能够成为一个 ip 段
* 判断的同时顺便把类型转了
*
* @param s
* @param left
* @param right
* @return
*/
private int judgeIfIpSegment(String s, int left, int right) {
int len = right - left + 1;
// 大于 1 位的时候，不能以 0 开头
if (len > 1 && s.charAt(left) == '0') {
return -1;
}
// 转成 int 类型
int res = 0;
for (int i = left; i <= right; i++) {
res = res * 10 + s.charAt(i) - '0';
}
if (res > 255) {
return -1;
}
return res;
}
private void dfs(String s, int len, int split, int begin, Deque<String> path, List<String> res) {
if (begin == len) {
if (split == 4) {
res.add(String.join(".", path));
}
return;
}
// 看到剩下的不够了，就退出（剪枝），len - begin 表示剩余的还未分割的字符串的位数
if (len - begin < (4 - split) || len - begin > 3 * (4 - split)) {
return;
}
for (int i = 0; i < 3; i++) {
if (begin + i >= len) {
break;
}
int ipSegment = judgeIfIpSegment(s, begin, begin + i);
if (ipSegment != -1) {
// 在判断是 ip 段的情况下，才去做截取
path.addLast(ipSegment + "");
dfs(s, len, split + 1, begin + i + 1, path, res);
path.removeLast();
}
}
}
}
子集
class Solution {
List<List<Integer>> result = new ArrayList<>();// 存放符合条件结果的集合
LinkedList<Integer> path = new LinkedList<>();// 用来存放符合条件结果
public List<List<Integer>> subsets(int[] nums) {
subsetsHelper(nums, 0);
return result;
}
private void subsetsHelper(int[] nums, int startIndex){
result.add(new ArrayList<>(path));//「遍历这个树的时候，把所有节点都记录下来，就是要求的子集集合」。
if (startIndex >= nums.length){ //终止条件可不加
return;
}
for (int i = startIndex; i < nums.length; i++){
path.add(nums[i]);
subsetsHelper(nums, i + 1);
path.removeLast();
}
}
}
子集II
注意去重 还是需要同一层上没有使用过
class Solution {
List<List<Integer>> result = new ArrayList<>();// 存放符合条件结果的集合
LinkedList<Integer> path = new LinkedList<>();// 用来存放符合条件结果
boolean[] used;
public List<List<Integer>> subsetsWithDup(int[] nums) {
if (nums.length == 0){
result.add(path);
return result;
}
Arrays.sort(nums);
used = new boolean[nums.length];
subsetsWithDupHelper(nums, 0);
return result;
}
private void subsetsWithDupHelper(int[] nums, int startIndex){
result.add(new ArrayList<>(path));
if (startIndex >= nums.length){
return;
}
for (int i = startIndex; i < nums.length; i++){
if (i > 0 && nums[i] == nums[i - 1] && !used[i - 1]){
continue;
}
path.add(nums[i]);
used[i] = true;
subsetsWithDupHelper(nums, i + 1);
path.removeLast();
used[i] = false;
}
}
}
递增子序列
注意本题不能 先对数组进行排序，排序后那么将全是递增子序列，比如对于数组[4,7,6,7]
对于used变量，在每一层进入时都会重置，所以回溯时不再需要重置！
package day.d0317;
import java.util.ArrayList;
import java.util.List;
/**
* @author hdbing
* @create 2023-03-17 15:18
*/
public class T02 {
public static void main(String[] args) {
Solution1 solution1 = new Solution1();
System.out.println(solution1.findSubsequences(new int[]{4, 6, 7, 6}));
}
}
class Solution1 {
private List<Integer> path = new ArrayList<>();
private List<List<Integer>> res = new ArrayList<>();
public List<List<Integer>> findSubsequences(int[] nums) {
backtracking(nums, 0);
return res;
}
private void backtracking(int[] nums, int start) {
if (path.size() > 1) {
res.add(new ArrayList<>(path));
}
int[] used = new int[201]; // 重置
for (int i = start; i < nums.length; i++) {
if (!path.isEmpty() && nums[i] < path.get(path.size() - 1) || (used[nums[i] + 100] == 1))
continue;
used[nums[i] + 100] = 1;
path.add(nums[i]);
backtracking(nums, i + 1);
path.remove(path.size() - 1);
}
}
}
全排列
class Solution {
List<List<Integer>> result = new ArrayList<>();// 存放符合条件结果的集合
LinkedList<Integer> path = new LinkedList<>();// 用来存放符合条件结果
boolean[] used;
public List<List<Integer>> permute(int[] nums) {
if (nums.length == 0){
return result;
}
used = new boolean[nums.length];
permuteHelper(nums);
return result;
}
private void permuteHelper(int[] nums){
if (path.size() == nums.length){
result.add(new ArrayList<>(path));
return;
}
for (int i = 0; i < nums.length; i++){
if (used[i]){
continue;
}
used[i] = true;
path.add(nums[i]);
permuteHelper(nums);
path.removeLast();
used[i] = false;
}
}
}
全排列 II
class Solution {
//存放结果
List<List<Integer>> result = new ArrayList<>();
//暂存结果
List<Integer> path = new ArrayList<>();
public List<List<Integer>> permuteUnique(int[] nums) {
boolean[] used = new boolean[nums.length];
Arrays.fill(used, false);
Arrays.sort(nums);
backTrack(nums, used);
return result;
}
private void backTrack(int[] nums, boolean[] used) {
if (path.size() == nums.length) {
result.add(new ArrayList<>(path));
return;
}
for (int i = 0; i < nums.length; i++) {
// used[i - 1] == true，说明同⼀树⽀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;
}
//如果同⼀树⽀nums[i]没使⽤过开始处理
if (used[i] == false) {
used[i] = true;//标记同⼀树⽀nums[i]使⽤过，防止同一树枝重复使用
path.add(nums[i]);
backTrack(nums, used);
path.remove(path.size() - 1);//回溯，说明同⼀树层nums[i]使⽤过，防止下一树层重复
used[i] = false;//回溯
}
}
}
}
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