General Approach for backtracking problem
来源:互联网 发布:000webhost绑定域名 编辑:程序博客网 时间:2024/06/05 16:04
https://leetcode.com/problems/combination-sum/discuss/
A general approach to backtracking questions in Java (Subsets, Permutations, Combination Sum, Palindrome Partitioning)
This structure might apply to many other backtracking questions, but here I am just going to demonstrate Subsets, Permutations, and Combination Sum.
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); 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++){ 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 : 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); } }}
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); 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++){ 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;}
- General Approach for backtracking problem
- A general approach to backtracking questions in Java (Subsets, Permutations, Combination Sum, Palind
- The BackTracking algorithm for n queen problem
- A New General Deep Learning Approach for Natural Language Processing
- Optimization of BackTracking algorithm for n queen problem
- Nathan A. Good, «Regular Expression Recipes for Windows Developers: A Problem-Solution Approach»
- RMAN Recipes for Oracle Database 11g: A Problem-Solution Approach
- backtracking
- BackTracking
- Backtracking
- Backtracking
- Backtracking
- Backtracking
- 【索引】General Problem Solving Techniques
- 【索引】General Problem Solving Techniques
- PHP 5 Recipes: A Problem-Solution Approach
- JDBC Recipes: A Problem-Solution Approach
- Shell Scripting Recipes: A Problem-Solution Approach
- prime算法求最小生成图
- java基本类型与包装类型
- Babun,一个开箱即用的 Windows Shell
- POJ1007 字符串按逆序对排序
- C基础之指针
- General Approach for backtracking problem
- Spring-AOP @AspectJ进阶之绑定类注解对象
- Leetcode--Add Two Numbers
- 哪种编程语言最好?
- BZOJ 1316: 树上的询问 点分治题解
- 李宏毅机器学习课程12~~~半监督学习
- Spring-AOP @AspectJ进阶之绑定连接点方法的返回值
- 【084】深度学习读书笔记:P26正交矩阵
- 什么是微积分?