标签:isp ++ ide 子集 splay 排列 else uniq can
1 Subsets
public List<List<Integer>> subsets(int[] nums) { List<List<Integer>> result = new ArrayList<>(); if (nums.length == 0) { return result; } ArrayList<Integer> list = new ArrayList<>(); subsetsHelper(nums, 0, list, result); return result; } void subsetsHelper(int[] nums, int pos, List<Integer> list, List<List<Integer>> result){ result.add(new ArrayList<>(list)); for (int i = pos; i < nums.length; i++) { list.add(nums[i]); subsetsHelper(nums, i + 1, list, result); list.remove(list.size() - 1); } }
2 Subsets II
public List<List<Integer>> subsetsWithDup(int[] nums) { List<List<Integer>> result = new ArrayList<>(); if (nums.length == 0) { return result; } Arrays.sort(nums); ArrayList<Integer> list = new ArrayList<>(); subsetsHelper(nums, 0, list, result); return result; } void subsetsHelper(int[] nums, int pos, List<Integer> list, List<List<Integer>> result){ result.add(new ArrayList<>(list)); for (int i = pos; i < nums.length; i++) { if (i > pos && nums[i] == nums[i - 1]){ continue; } list.add(nums[i]); subsetsHelper(nums, i + 1, list, result); list.remove(list.size() - 1); } }
3 Permutations
public List<List<Integer>> permute(int[] nums) { // write your code here List<List<Integer>> res = new ArrayList<>(); if (nums == null || nums.length == 0) { res.add(new ArrayList<Integer>()); return res; } List<Integer> list = new ArrayList<>(); helper(nums, list, res); return res; } void helper(int[] nums, List<Integer> list, List<List<Integer>> res) { if (list.size() == nums.length) { res.add(new ArrayList<>(list)); return; } for (int i = 0; i < nums.length; i++) { if (list.contains(nums[i])) { continue; } list.add(nums[i]); helper(nums, list, res); list.remove(list.size() - 1); } }
4 Permutations II
public List<List<Integer>> permuteUnique(int[] nums) { // Write your code here List<List<Integer>> result = new ArrayList<>(); if (nums == null || nums.length == 0) { result.add(new ArrayList<Integer>()); return result; } Arrays.sort(nums); List<Integer> list = new ArrayList<>(); boolean[] used = new boolean[nums.length]; helper(nums, list, used, result); return result; } void helper(int[] nums, List<Integer> list, boolean[] used, List<List<Integer>> result) { if (list.size() == nums.length) { result.add(new ArrayList<>(list)); return; } 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; list.add(nums[i]); helper(nums, list, used, result); used[i] = false; list.remove(list.size() - 1); } }
5 Combination Sum 可以有重复
public ArrayList<ArrayList<Integer>> combinationSum(int[] candidates, int target) { ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>(); if (candidates == null || candidates.length == 0) { return res; } Arrays.sort(candidates); helper(candidates, 0, target, new ArrayList<Integer>(), res); return res; } private void helper(int[] candidates, int start, int target, ArrayList<Integer> item, ArrayList<ArrayList<Integer>> res) { // TODO Auto-generated method stub if (target < 0) { return; } if (target == 0) { res.add(new ArrayList<>(item)); return; } for (int i = start; i < candidates.length; i++) { if (i > 0 && candidates[i] == candidates[i - 1]) { continue; } item.add(candidates[i]); helper(candidates, i, target - candidates[i], item, res); item.remove(item.size() - 1); } }
6 Combination Sum 不可以有重复
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); } } }
7 Palindrome Partitioning
public class Solution { public List<List<String>> partition(String s) { List<List<String>> res = new ArrayList<>(); if (s.length() == 0) return res; dfs(s, 0, new ArrayList<String>(), res); return res; } public void dfs(String s, int index, ArrayList<String> item, List<List<String>> res){ if (s.length() == index) { res.add(new ArrayList<>(item)); return; } for (int i = index; i < s.length(); i++){ if (isP(s, index, i)){ item.add(s.substring(index, i + 1)); dfs(s, i + 1, item, res); item.remove(item.size() - 1); } } } public boolean isP(String s, int l, int r){ while (l < r) { if (s.charAt(l) != s.charAt(r)){ return false; } l++;r--; } return true; } }
标签:isp ++ ide 子集 splay 排列 else uniq can
原文地址:http://www.cnblogs.com/whesuanfa/p/7435894.html