标签:splay dba amp 数字 start isp i+1 spl hat
找出所有方案
排列和组合问题
排列:
https://www.lintcode.com/problem/combination-sum/description
1 public class Solution { 2 /** 3 * @param candidates: A list of integers 4 * @param target: An integer 5 * @return: A list of lists of integers 6 */ 7 public List<List<Integer>> combinationSum(int[] candidates, int target) { 8 // write your code here 9 List<List<Integer>> results = new ArrayList<>(); 10 if(candidates == null){ 11 return results; 12 } 13 Arrays.sort(candidates); 14 15 List<Integer> combination = new ArrayList<>(); 16 helper(candidates,0,results,target,combination); 17 18 return results; 19 } 20 21 //1.定义:找到所有combination开头的组合,后面的和是target的组合 22 private void helper(int[] candidates, 23 int startIndex, List<List<Integer>> results, 24 int target,List<Integer> combination){ 25 //3.出口 26 if(target==0){ 27 results.add(new ArrayList<Integer>(combination)); 28 return; 29 } 30 //2.拆解 31 for(int i = startIndex;i<candidates.length;i++){ 32 if(target<candidates[i]){ 33 break; 34 } 35 36 if(i-1>=0 && candidates[i]==candidates[i-1]){ 37 continue; 38 } 39 40 combination.add(candidates[i]); 41 helper(candidates,i,results,target-candidates[i],combination); 42 combination.remove(combination.size()-1); 43 } 44 } 45 }
https://www.lintcode.com/problem/combination-sum-ii/description
1 public class Solution { 2 /** 3 * @param num: Given the candidate numbers 4 * @param target: Given the target number 5 * @return: All the combinations that sum to target 6 */ 7 public List<List<Integer>> combinationSum2(int[] num, int target) { 8 // write your code here 9 List<List<Integer>> results = new ArrayList<>(); 10 if(num == null){ 11 return results; 12 } 13 14 Arrays.sort(num); 15 16 List<Integer> combination = new ArrayList<>(); 17 helper(num,0,results,target,combination); 18 19 return results; 20 } 21 22 private void helper(int[]num,int startIndex,List<List<Integer>> results, 23 int target,List<Integer> combination){ 24 if(target == 0){ 25 results.add(new ArrayList<Integer>(combination)); 26 return; 27 } 28 29 for(int i = startIndex;i<num.length;i++){ 30 if(target<num[i]){ 31 break; 32 } 33 34 if(i-1>=0 && i!=startIndex && num[i]==num[i-1]){ 35 continue; 36 } 37 38 combination.add(num[i]); 39 helper(num,i+1,results,target-num[i],combination); 40 combination.remove(combination.size()-1); 41 } 42 } 43 44 }
切割问题
N个字母的字符串对应N-1个数字的组合
N个字符的字符串 子串:O(N^2)个
排列:
https://www.lintcode.com/problem/permutations/description
1 public class Solution { 2 /* 3 * @param nums: A list of integers. 4 * @return: A list of permutations. 5 */ 6 public List<List<Integer>> permute(int[] nums) { 7 // write your code here 8 List<List<Integer>> rst = new ArrayList<>(); 9 if(nums==null){ 10 return rst; 11 } 12 13 if(nums.length ==0){ 14 rst.add(new ArrayList<>()); 15 return rst; 16 } 17 18 List<Integer> list = new ArrayList<>(); 19 helper(rst,list,nums); 20 return rst; 21 } 22 23 private void helper(List<List<Integer>> rst, List<Integer> list, int[] nums){ 24 if(list.size()==nums.length){ 25 rst.add(new ArrayList<Integer>(list)); 26 return; 27 } 28 29 for(int i=0;i<nums.length;i++){ 30 if(list.contains(nums[i])){ 31 continue; 32 } 33 34 list.add(nums[i]); 35 helper(rst,list,nums); 36 list.remove(list.size()-1); 37 } 38 } 39 }
https://www.lintcode.com/problem/permutations-ii/description
1 public class Solution { 2 /* 3 * @param : A list of integers 4 * @return: A list of unique permutations 5 */ 6 public List<List<Integer>> permuteUnique(int[] nums) { 7 // write your code here 8 List<List<Integer>> rst = new ArrayList<>(); 9 if(nums==null){ 10 return rst; 11 } 12 13 if(nums.length ==0){ 14 rst.add(new ArrayList<>()); 15 return rst; 16 } 17 18 Arrays.sort(nums); 19 List<Integer> list = new ArrayList<>(); 20 int[] visited = new int[nums.length]; 21 helper(rst,list,nums,visited); 22 return rst; 23 } 24 25 private void helper(List<List<Integer>> rst, List<Integer> list, int[] nums,int[] visited){ 26 if(list.size()==nums.length){ 27 rst.add(new ArrayList<Integer>(list)); 28 return; 29 } 30 31 for(int i=0;i<nums.length;i++){ 32 if(visited[i]==1){ 33 continue; 34 } 35 if(i-1>=0 && nums[i]==nums[i-1] && visited[i-1]==0){ 36 continue; 37 } 38 39 list.add(nums[i]); 40 visited[i]=1; 41 42 helper(rst,list,nums,visited); 43 44 list.remove(list.size()-1); 45 visited[i]=0; 46 } 47 } 48 };
DFS 时间复杂度:答案个数*构造每个答案的时间
DP:状态个数*计算每个状态的时间
标签:splay dba amp 数字 start isp i+1 spl hat
原文地址:https://www.cnblogs.com/lizzyluvcoding/p/10354366.html