Combination Sum | & || & ||| & IV

Combination Sum |

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

The same repeated number may be chosen from C unlimited number of times.

For example, given candidate set 2,3,6,7 and target 7, 
A solution set is: 
[7] 
[2, 2, 3]

Notice

• 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.

 

Example

given candidate set 2,3,6,7 and target 7, 
A solution set is: 
[7] 
[2, 2, 3]

分析：递归

 public class Solution {
     public List<List<Integer>> combinationSum(int[] candidates, int target) {
         List<List<Integer>> listsAll = new ArrayList<>();
         Arrays.sort(candidates);
         helper(, , candidates, target, new ArrayList<>(), listsAll);
         return listsAll;
     }

     public static void helper(int index, int total, int[] candidates, int target, List<Integer> list, List<List<Integer>> listsAll) {
         if (index >= candidates.length || total >= target) return;
         list.add(candidates[index]);
         total += candidates[index];
         if (total == target) {
             listsAll.add(new ArrayList<>(list));
         }
         helper(index, total, candidates, target, list, listsAll);
         total = total - candidates[index];
         list.remove(list.size() - );
         helper(index + , total, candidates, target, list, listsAll);
     }
 }

Combination Sum II

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.

Notice

• 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.

Example

Given candidate set [10,1,6,7,2,1,5] and target 8,

A solution set is:

[
  [1,7],
  [1,2,5],
  [2,6],
  [1,1,6]
]

 public class Solution {
     public List<List<Integer>> combinationSum2(int[] candidates, int target) {
         List<List<Integer>> listsAll = new ArrayList<List<Integer>>();
         Arrays.sort(candidates);
         helper(, , candidates, target, new ArrayList<>(), listsAll);
         return listsAll;
     }

     public static void helper(int index, int total, int[] candidates, int target, List<Integer> list, List<List<Integer>> listsAll) {
         if (index >= candidates.length || total >= target) return;
         list.add(candidates[index]);
         total += candidates[index];
         if (total == target) {
             listsAll.add(new ArrayList<Integer>(list));
         }
         helper(index + , total, candidates, target, list, listsAll);
         total = total - candidates[index];
         list.remove(list.size() - );
         while (index +  < candidates.length && candidates[index] == candidates[index + ]) {
             index++;
         }
         helper(index + , total, candidates, target, list, listsAll);
     }
 }


Combination Sum IV

Given an integer array with all positive numbers and no duplicates, find the number of possible combinations that add up to a positive integer target.

Example:

nums = [1, 2, 3]
target = 4

The possible combination ways are:
(1, 1, 1, 1)
(1, 1, 2)
(1, 2, 1)
(1, 3)
(2, 1, 1)
(2, 2)
(3, 1)

Note that different sequences are counted as different combinations.

Therefore the output is 7.
分析： 这题和change coin非常相似。

 public class Solution {
     public int combinationSum4(int[] nums, int target) {
         if (nums == null || nums.length == ) return ;

         int[] dp = new int[target + ];
         dp[] = ;

         for (int i = ; i <= target; i++) {
             for (int num : nums) {
                 if (i - num >= ) {
                     dp[i] += dp[i - num];
                 }
             }
         }
         return dp[target];
     }
 }

参考请注明出处：cnblogs.com/beiyeqingteng/