package 算法和数据结构;
/** * Filename : Backtracking.java * Author : zhihao_tian@126.com * Creation time : 上午10:16:04 - 2017年3月13日 * Description : 利用回溯法来解决诸如 子集数量,排列,组合 的问题。 */
import java.util.*;
public class Backtracking {
/** * Subsets */
//**********************************************************************
//Subsets : https://leetcode.com/problems/subsets/
public List<List<Integer>> subsets(int[] nums) {
List<List<Integer>> list = new ArrayList<>();
Arrays.sort(nums);
backtrack(list, new ArrayList<>(), nums, 0);
return list;
}
private void backtrack(List<List<Integer>> list , List<Integer> tempList, int [] nums, int start){
list.add(new ArrayList<>(tempList));
for(int i = start; i < nums.length; i++){
tempList.add(nums[i]);
backtrack(list, tempList, nums, i + 1);
tempList.remove(tempList.size() - 1);
}
}
//********************************************************************
//Subsets II (contains duplicates) : https://leetcode.com/problems/subsets-ii/
public List<List<Integer>> subsetsWithDup(int[] nums) {
List<List<Integer>> list = new ArrayList<>();
Arrays.sort(nums);
backtrack2(list, new ArrayList<>(), nums, 0);
return list;
}
private void backtrack2(List<List<Integer>> list, List<Integer> tempList, int [] nums, int start){
list.add(new ArrayList<>(tempList));
for(int i = start; i < nums.length; i++){
if(i > start && nums[i] == nums[i-1]) continue; // skip duplicates
tempList.add(nums[i]);
backtrack(list, tempList, nums, i + 1);
tempList.remove(tempList.size() - 1);
}
}
/** * Permutations排列 */
//***************************************************************************
//Permutations : https://leetcode.com/problems/permutations/
public List<List<Integer>> permute(int[] nums) {
List<List<Integer>> list = new ArrayList<>();
// Arrays.sort(nums); // not necessary
backtrack(list, new ArrayList<>(), nums);
return list;
}
private void backtrack(List<List<Integer>> list, List<Integer> tempList, int [] nums){
if(tempList.size() == nums.length){
list.add(new ArrayList<>(tempList));
} else{
for(int i = 0; i < nums.length; i++){
if(tempList.contains(nums[i])) continue; // element already exists, skip
tempList.add(nums[i]);
backtrack(list, tempList, nums);
tempList.remove(tempList.size() - 1);
}
}
}
//*******************************************************************************************
//Permutations II (contains duplicates) : https://leetcode.com/problems/permutations-ii/
public List<List<Integer>> permuteUnique(int[] nums) {
List<List<Integer>> list = new ArrayList<>();
Arrays.sort(nums);
backtrack(list, new ArrayList<>(), nums, new boolean[nums.length]);
return list;
}
private void backtrack(List<List<Integer>> list, List<Integer> tempList, int [] nums, boolean [] used) {
if(tempList.size() == nums.length){
list.add(new ArrayList<>(tempList));
} else {
for(int i = 0; i < nums.length; i++){
if(used[i] || i > 0 && nums[i] == nums[i-1] && !used[i - 1]) continue;
used[i] = true;
tempList.add(nums[i]);
backtrack(list, tempList, nums, used);
used[i] = false;
tempList.remove(tempList.size() - 1);
}
}
}
/** * 组合 */
//****************************************************************************
//Combination Sum : https://leetcode.com/problems/combination-sum/
public List<List<Integer>> combinationSum(int[] nums, int target) {
List<List<Integer>> list = new ArrayList<>();
Arrays.sort(nums);
backtrack(list, new ArrayList<>(), nums, target, 0);
return list;
}
private void backtrack(List<List<Integer>> list, List<Integer> tempList, int [] nums, int remain, int start){
if(remain < 0) return;
else if(remain == 0) list.add(new ArrayList<>(tempList));
else{
for(int i = start; i < nums.length; i++){
tempList.add(nums[i]);
backtrack(list, tempList, nums, remain - nums[i], i); // not i + 1 because we can reuse same elements
tempList.remove(tempList.size() - 1);
}
}
}
//Combination Sum II (can't reuse same element) : https://leetcode.com/problems/combination-sum-ii/
public List<List<Integer>> combinationSum2(int[] nums, int target) {
List<List<Integer>> list = new ArrayList<>();
Arrays.sort(nums);
backtrack2(list, new ArrayList<>(), nums, target, 0);
return list;
}
private void backtrack2(List<List<Integer>> list, List<Integer> tempList, int [] nums, int remain, int start){
if(remain < 0) return;
else if(remain == 0) list.add(new ArrayList<>(tempList));
else{
for(int i = start; i < nums.length; i++){
if(i > start && nums[i] == nums[i-1]) continue; // skip duplicates
tempList.add(nums[i]);
backtrack(list, tempList, nums, remain - nums[i], i + 1);
tempList.remove(tempList.size() - 1);
}
}
}
//Palindrome Partitioning : https://leetcode.com/problems/palindrome-partitioning/
public List<List<String>> partition(String s) {
List<List<String>> list = new ArrayList<>();
backtrack(list, new ArrayList<>(), s, 0);
return list;
}
public void backtrack(List<List<String>> list, List<String> tempList, String s, int start){
if(start == s.length())
list.add(new ArrayList<>(tempList));
else{
for(int i = start; i < s.length(); i++){
if(isPalindrome(s, start, i)){
tempList.add(s.substring(start, i + 1));
backtrack(list, tempList, s, i + 1);
tempList.remove(tempList.size() - 1);
}
}
}
}
public boolean isPalindrome(String s, int low, int high){
while(low < high)
if(s.charAt(low++) != s.charAt(high--)) return false;
return true;
}
}
回溯法解决排列组合问题
原文作者:回溯法
原文地址: https://blog.csdn.net/sdu_bupt/article/details/71512956
本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
原文地址: https://blog.csdn.net/sdu_bupt/article/details/71512956
本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。