377. 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.

Follow up:
What if negative numbers are allowed in the given array?
How does it change the problem?
What limitation we need to add to the question to allow negative numbers?

本题解法我一开始以为是用回溯法，后来发现TLE了，看了答案知道是DP

DP是将原问题分解成若干规模较小的子问题，原问题是求target的组成元素个数和，则子问题可以是1～target的各个组成之和，dp数组里面保存就是小于target的和的组成个数。代码如下：

 1 public class Solution {
 2     public int combinationSum4(int[] nums, int target) {
 3         int[] dp = new int[target+1];
 4         Arrays.sort(nums);
 5         for(int i=1;i<=target;i++){
 6             for(int num:nums){
 7                 if(i>num) dp[i] +=dp[i-num];
 8                 else if(i==num) dp[i]+=1;
 9                 else if(i<num) break;
10             }
11         }
12         return dp[target];
13     }
14 }

377. Combination Sum IV

原文：http://www.cnblogs.com/codeskiller/p/6389022.html