【Leetcode】Combination sum 1，2

【题目】

【分析】

首先，先sort,

如果target > 0:

【代码】

using recursive 这恐怕的 O(n2)...

过程：

before recursion : [2]

before recursion : [2, 2]

before recursion : [2, 2, 2]

after recursion and remove : [2, 2]

before recursion : [2, 2, 3]

after recursion and remove : [2, 2]

after recursion and remove : [2]

before recursion : [2, 3]

after recursion and remove : [2]

after recursion and remove : []

before recursion : [3]

before recursion : [3, 3]

after recursion and remove : [3]

after recursion and remove : []

before recursion : [6]

after recursion and remove : []

before recursion : [7]

after recursion and remove : []

[[2, 2, 3], [7]]

public class Solution {    public List<List<Integer>> combinationSum(int[] candidates, int target) {        Arrays.sort(candidates);        List<List<Integer>> result = new ArrayList<List<Integer>>();        getResult(result, new ArrayList<Integer>(), candidates, target, 0);        return result;    }    private void getResult(List<List<Integer>> result, List<Integer> cur, int candidates[], int target, int start){        if(target > 0){            for(int i = start; i < candidates.length && target >= candidates[i]; i++){                cur.add(candidates[i]);                getResult(result, cur, candidates, target - candidates[i], i);                cur.remove(cur.size() - 1);            }//for        }//if        else if(target == 0 ){            result.add(new ArrayList<Integer>(cur));        }//else if    }}

【二】

【题目】

Given a collection of candidate numbers (C) and a target number (T)，find all unique combinations in C where the candidate numbers sums to T.
Each number in C may only be used once in the combination.
Note:
All numbers (including target) will be positive integers.
Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
The solution set must not contain duplicate combinations.
For example, given candidate set 10,1,2,7,6,1,5 and target 8, 
A solution set is: 
[1, 7] 
[1, 2, 5] 
[2, 6] 
[1, 1, 6] 

【题目】

题意分析：从给定数组中找到一组数字，要求这组数字之和等于target。另外，数组中的数字不允许被使用多次，但如果一开始就存在多个的话，可以使用多次。
解题思路：显然先排序，然后dfs。其中有一点要注意的是：因为不能重复，所以要跳过一样的数字。以上面为例，如果不跳过重复的1的话，会出现多个：[1,7]

    public static void dfs(int [] num, int start, int target, List<Integer> array, List<List<Integer>> result) {          if(target==0) {              result.add(new ArrayList<Integer>(array));              return;          }                    if(start>=num.length||num[0]>target) {              return;          }          int i = start;          while(i<num.length) {              if(num[i]<=target) {                  array.add(num[i]);  System.out.println("before dfs: " + array);                dfs(num, i + 1, target - num[i], array, result);                  array.remove(array.size()-1);  System.out.println("after dfs and rempve: " + array);                //跳过重复的元素                  while(i<(num.length-1)&&num[i]==num[i+1]) {                      i++;                  }              }              i++;          }      }                  public static List<List<Integer>> combinationSum2(int[] num, int target) {           List<List<Integer>> result = new ArrayList<List<Integer>>();           ArrayList<Integer> array = new ArrayList<Integer>();           if(num==null) {               result.add(array);               return result;           }           Arrays.sort(num);           dfs(num,0, target,array,result);            return result;      }