DFS-20190206

找出所有方案

排列和组合问题

排列：

https://www.lintcode.com/problem/combination-sum/description

 public class Solution {
     /**
      * @param candidates: A list of integers
      * @param target: An integer
      * @return: A list of lists of integers
      */
     public List<List<Integer>> combinationSum(int[] candidates, int target) {
         // write your code here
         List<List<Integer>> results = new ArrayList<>();
         if(candidates == null){
             return results;
         }
         Arrays.sort(candidates);

         List<Integer> combination = new ArrayList<>();
         helper(candidates,0,results,target,combination);

         return results;
     }

     //1.定义：找到所有combination开头的组合，后面的和是target的组合
     private void helper(int[] candidates,
         int startIndex, List<List<Integer>> results,
         int target,List<Integer> combination){
             //3.出口
             if(target==0){
                 results.add(new ArrayList<Integer>(combination));
                 return;
             }
             //2.拆解
             for(int i = startIndex;i<candidates.length;i++){
                 if(target<candidates[i]){
                     break;
                 }

                 if(i-1>=0 && candidates[i]==candidates[i-1]){
                     continue;
                 }

                 combination.add(candidates[i]);
                 helper(candidates,i,results,target-candidates[i],combination);
                 combination.remove(combination.size()-1);
             }
         }
 }

https://www.lintcode.com/problem/combination-sum-ii/description

 public class Solution {
     /**
      * @param num: Given the candidate numbers
      * @param target: Given the target number
      * @return: All the combinations that sum to target
      */
     public List<List<Integer>> combinationSum2(int[] num, int target) {
         // write your code here
         List<List<Integer>> results = new ArrayList<>();
         if(num == null){
             return results;
         }

         Arrays.sort(num);

         List<Integer> combination = new ArrayList<>();
         helper(num,0,results,target,combination);

         return results;
     }

     private void helper(int[]num,int startIndex,List<List<Integer>> results,
             int target,List<Integer> combination){
             if(target == 0){
                 results.add(new ArrayList<Integer>(combination));
                 return;
             }

             for(int i = startIndex;i<num.length;i++){
                 if(target<num[i]){
                     break;
                 }

                 if(i-1>=0 && i!=startIndex && num[i]==num[i-1]){
                     continue;
                 }

                 combination.add(num[i]);
                 helper(num,i+1,results,target-num[i],combination);
                 combination.remove(combination.size()-1);
             }
         }

 }

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

 public class Solution {
     /*
      * @param nums: A list of integers.
      * @return: A list of permutations.
      */
     public List<List<Integer>> permute(int[] nums) {
         // write your code here
         List<List<Integer>> rst = new ArrayList<>();
         if(nums==null){
             return rst;
         }

         if(nums.length ==0){
             rst.add(new ArrayList<>());
             return rst;
         }

         List<Integer> list = new ArrayList<>();
         helper(rst,list,nums);
         return rst;
     }

     private void helper(List<List<Integer>> rst, List<Integer> list, int[] nums){
         if(list.size()==nums.length){
             rst.add(new ArrayList<Integer>(list));
             return;
         }

         for(int i=0;i<nums.length;i++){
             if(list.contains(nums[i])){
                 continue;
             }

             list.add(nums[i]);
             helper(rst,list,nums);
             list.remove(list.size()-1);
         }
     }
 }

https://www.lintcode.com/problem/permutations-ii/description

 public class Solution {
     /*
      * @param :  A list of integers
      * @return: A list of unique permutations
      */
     public List<List<Integer>> permuteUnique(int[] nums) {
         // write your code here
         List<List<Integer>> rst = new ArrayList<>();
         if(nums==null){
             return rst;
         }

         if(nums.length ==0){
             rst.add(new ArrayList<>());
             return rst;
         }

         Arrays.sort(nums);
         List<Integer> list = new ArrayList<>();
         int[] visited = new int[nums.length];
         helper(rst,list,nums,visited);
         return rst;
     }

         private void helper(List<List<Integer>> rst, List<Integer> list, int[] nums,int[] visited){
         if(list.size()==nums.length){
             rst.add(new ArrayList<Integer>(list));
             return;
         }

         for(int i=0;i<nums.length;i++){
             if(visited[i]==1){
                 continue;
             }
             if(i-1>=0 && nums[i]==nums[i-1] && visited[i-1]==0){
                 continue;
             }

             list.add(nums[i]);
             visited[i]=1;

             helper(rst,list,nums,visited);

             list.remove(list.size()-1);
             visited[i]=0;
         }
     }
 };

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